MUST-HAVE MATH TOOLS FOR GRADUATE STUDY IN ECONOMICS

MUST-HAVE MATH TOOLS FOR GRADUATE STUDY IN ECONOMICS William Neilson Department of Economics University of Tennessee – Knoxville September 2009 © 2008-9 by William Neilson web.utk.edu/~wneilson/mathbook.pdf Acknowledgments Valentina Kozlova, Kelly Padden, and John Tilstra provided valuable proofreading assistance on the first version of this book, and I am grateful. Other mistakes were found by the students in my class. Of course, if they missed anything it is still my fault. Valentina and Bruno Wichmann have both suggested additions to the book, including the sections on stability of dynamic systems and order statistics. The cover picture was provided by my son, Henry, who also proofread parts of the book. I have always liked this picture, and I thank him for letting me use it. CONTENTS 1 Econ and math 1.1 Some important graphs . . . . . . . . . . . . . . . . . . . . . . 1.2 Math, micro, and metrics . . . . . . . . . . . . . . . . . . . . . 1 2 4 I 6 Optimization (Multivariate calculus) 2 Single variable optimization 2.1 A graphical approach . . . . . . . . . . . 2.2 Derivatives . . . . . . . . . . . . . . . . . 2.3 Uses of derivatives . . . . . . . . . . . . 2.4 Maximum or minimum? . . . . . . . . . 2.5 Logarithms and the exponential function 2.6 Problems . . . . . . . . . . . . . . . . . . 3 Optimization with several variables 3.1 A more complicated pro…t function 3.2 Vectors and Euclidean space . . . . 3.3 Partial derivatives . . . . . . . . . . 3.4 Multidimensional optimization . . . i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 8 9 14 16 17 18 . . . . 21 21 22 24 26 ii 3.5 Comparative statics analysis . . . . . . . . . . . . . . . . . . . 29 3.5.1 An alternative approach (that I don’t like) . . . . . . . 31 3.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 4 Constrained optimization 4.1 A graphical approach . . . . . . . . . 4.2 Lagrangians . . . . . . . . . . . . . . 4.3 A 2-dimensional example . . . . . . . 4.4 Interpreting the Lagrange multiplier . 4.5 A useful example - Cobb-Douglas . . 4.6 Problems . . . . . . . . . . . . . . . . 5 Inequality constraints 5.1 Lame example - capacity constraints 5.1.1 A binding constraint . . . . . 5.1.2 A nonbinding constraint . . . 5.2 A new approach . . . . . . . . . . . . 5.3 Multiple inequality constraints . . . . 5.4 A linear programming example . . . 5.5 Kuhn-Tucker conditions . . . . . . . 5.6 Problems . . . . . . . . . . . . . . . . II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 37 39 40 42 43 48 . . . . . . . . 52 53 54 55 56 59 62 64 67 Solving systems of equations (Linear algebra) 6 Matrices 6.1 Matrix algebra . . 6.2 Uses of matrices . . 6.3 Determinants . . . 6.4 Cramer’s rule . . . 6.5 Inverses of matrices 6.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Systems of equations 7.1 Identifying the number of solutions 7.1.1 The inverse approach . . . . 7.1.2 Row-echelon decomposition 7.1.3 Graphing in (x,y) space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 . . . . . . . . . . . . . . . . 72 72 76 77 79 81 83 . . . . 86 87 87 87 89 iii 7.1.4 Graphing in column space . . . . . . . . . . . . . . . . 89 7.2 Summary of results . . . . . . . . . . . . . . . . . . . . . . . . 91 7.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 8 Using linear algebra in economics 8.1 IS-LM analysis . . . . . . . . . . . . 8.2 Econometrics . . . . . . . . . . . . . 8.2.1 Least squares analysis . . . . 8.2.2 A lame example . . . . . . . . 8.2.3 Graphing in column space . . 8.2.4 Interpreting some matrices . 8.3 Stability of dynamic systems . . . . . 8.3.1 Stability with a single variable 8.3.2 Stability with two variables . 8.3.3 Eigenvalues and eigenvectors . 8.3.4 Back to the dynamic system . 8.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Second-order conditions 9.1 Taylor approximations for R ! R . . . . . . . 9.2 Second order conditions for R ! R . . . . . . 9.3 Taylor approximations for Rm ! R . . . . . . 9.4 Second order conditions for Rm ! R . . . . . 9.5 Negative semide…nite matrices . . . . . . . . . 9.5.1 Application to second-order conditions 9.5.2 Examples . . . . . . . . . . . . . . . . 9.6 Concave and convex functions . . . . . . . . . 9.7 Quasiconcave and quasiconvex functions . . . 9.8 Problems . . . . . . . . . . . . . . . . . . . . . III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 . 114 . 116 . 116 . 118 . 118 . 119 . 120 . 120 . 124 . 128 Econometrics (Probability and statistics) 10 Probability 10.1 Some de…nitions . . . . . . . . . 10.2 De…ning probability abstractly . 10.3 De…ning probabilities concretely 10.4 Conditional probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 95 98 98 99 100 101 102 102 104 105 108 109 130 . . . . . . . . . . . . . . . . 131 . 131 . 132 . 134 . 136 iv 10.5 10.6 10.7 10.8 Bayes’ rule . . . . . . . . Monty Hall problem . . Statistical independence Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 139 140 141 11 Random variables 11.1 Random variables . . . . . . . . . . . . . . 11.2 Distribution functions . . . . . . . . . . . 11.3 Density functions . . . . . . . . . . . . . . 11.4 Useful distributions . . . . . . . . . . . . . 11.4.1 Binomial (or Bernoulli) distribution 11.4.2 Uniform distribution . . . . . . . . 11.4.3 Normal (or Gaussian) distribution . 11.4.4 Exponential distribution . . . . . . 11.4.5 Lognormal distribution . . . . . . . 11.4.6 Logistic distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 143 144 144 145 145 147 148 149 151 151 12 Integration 12.1 Interpreting integrals . . . . . . . . . . . . . . 12.2 Integration by parts . . . . . . . . . . . . . . . 12.2.1 Application: Choice between lotteries 12.3 Di¤erentiating integrals . . . . . . . . . . . . . 12.3.1 Application: Second-price auctions . . 12.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 . 155 . 156 . 157 . 159 . 161 . 163 13 Moments 13.1 Mathematical expectation . 13.2 The mean . . . . . . . . . . 13.2.1 Uniform distribution 13.2.2 Normal distribution . 13.3 Variance . . . . . . . . . . . 13.3.1 Uniform distribution 13.3.2 Normal distribution . 13.4 Application: Order statistics 13.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 164 165 165 165 166 167 168 168 173 v 14 Multivariate distributions 14.1 Bivariate distributions . . . . . . . . . . . . . . . . . . . . . 14.2 Marginal and conditional densities . . . . . . . . . . . . . . . 14.3 Expectations . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Conditional expectations . . . . . . . . . . . . . . . . . . . . 14.4.1 Using conditional expectations - calculating the bene…t of search . . . . . . . . . . . . . . . . . . . . . . . . . 14.4.2 The Law of Iterated Expectations . . . . . . . . . . . 14.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 . 175 . 176 . 178 . 181 . 181 . 184 . 185 15 Statistics 15.1 Some de…nitions . . . . . . . . . . 15.2 Sample mean . . . . . . . . . . . 15.3 Sample variance . . . . . . . . . . 15.4 Convergence of random variables 15.4.1 Law of Large Numbers . . 15.4.2 Central Limit Theorem . . 15.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 187 188 189 192 192 193 193 16 Sampling distributions 16.1 Chi-square distribution . . . . . . . . . . 16.2 Sampling from the normal distribution . 16.3 t and F distributions . . . . . . . . . . . 16.4 Sampling from the binomial distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 194 196 198 200 17 Hypothesis testing 17.1 Structure of hypothesis tests . . 17.2 One-tailed and two-tailed tests . 17.3 Examples . . . . . . . . . . . . 17.3.1 Example 1 . . . . . . . . 17.3.2 Example 2 . . . . . . . . 17.3.3 Example 3 . . . . . . . . 17.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 202 205 207 207 208 208 209 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Solutions to end-of-chapter problems 211 18.1 Solutions for Chapter 15 . . . . . . . . . . . . . . . . . . . . . 265 Index 267 CHAPTER 1 Econ and math Every academic discipline has its own standards by which it judges the merits of what researchers claim to be true. In the physical sciences this typically requires experimental veri…cation. In history it requires links to the original sources. In sociology one can often get by with anecdotal evidence, that is, with giving examples. In economics there are two primary ways one can justify an assertion, either using empirical evidence (econometrics or experimental work) or mathematical arguments. Both of these techniques require some math, and one purpose of this course is to provide you with the mathematical tools needed to make and understand economic arguments. A second goal, though, is to teach you to speak mathematics as a second language, that is, to make you comfortable talking about economics using the shorthand of mathematics. In undergraduate courses economic arguments are often made using graphs. In graduate courses we tend to use equations. But equations often have graphical counterparts and vice versa. Part of getting comfortable about using math to do economics is knowing how to go from graphs to the underlying equations, and part is going from equations to the appropriate graphs. 1 CHAPTER 1. ECON AND MATH 2 Figure 1.1: A constrained choice problem 1.1 Some important graphs One of the fundamental graphs is shown in Figure 1.1. The axes and curves are not labeled, but that just ampli…es its importance. If the axes are commodities, the line is a budget line, and the curve is an indi¤erence curve, the graph depicts the fundamental consumer choice problem. If the axes are inputs, the curve is an isoquant, and the line is an iso-cost line, the graph illustrates the …rm’s cost-minimization problem. Figure 1.1 raises several issues. How do we write the equations for the line and the curve? The line and curve seem to be tangent. How do we characterize tangency? At an even more basic level, how do we …nd slopes of curves? How do we write conditions for the curve to be curved the way it is? And how do we do all of this with equations instead of a picture? Figure 1.2 depicts a di¤erent situation. If the upward-sloping line is a supply curve and the downward-sloping one is a demand curve, the graph shows how the market price is determined. If the upward-sloping line is marginal cost and the downward-sloping line is marginal bene…t, the …gure shows how an individual or …rm chooses an amount of some activity. The questions for Figure 1.2 are: How do we …nd the point where the two lines intersect? How do we …nd the change from one intersection point to another? And how do we know that two curves will intersect in the …rst place? Figure 1.3 is completely di¤erent. It shows a collection of points with a CHAPTER 1. ECON AND MATH 3 Figure 1.2: Solving simultaneous equations line …tting through them. How do we …t the best line through these points? This is the key to doing empirical work. For example, if the horizontal axis measures the quantity of a good and the vertical axis measures its price, the points could be observations of a demand curve. How do we …nd the demand curve that best …ts the data? These three graphs are fundamental to economics. There are more as well. All of them, though, require that we restrict attention to two dimensions. For the …rst graph that means consumer choice with only two commodities, but we might want to talk about more. For the second graph it means supply and demand for one commodity, but we might want to consider several markets simultaneously. The third graph allows quantity demanded to depend on price, but not on income, prices of other goods, or any other factors. So, an important question, and a primary reason for using equations instead of graphs, is how do we handle more than two dimensions? Math does more for us than just allow us to expand the number of dimensions. It provides rigor; that is, it allows us to make sure that our statements are true. All of our assertions will be logical conclusions from our initial assumptions, and so we know that our arguments are correct and we can then devote attention to the quality of the assumptions underlying them. CHAPTER 1. ECON AND MATH 4 Figure 1.3: Fitting a line to data points 1.2 Math, micro, and metrics The theory of microeconomics is based on two primary concepts: optimization and equilibrium. Finding how much a …rm produces to maximize pro…t is an example of an optimization problem, as is …nding what a consumer purchases to maximize utility. Optimization problems usually require …nding maxima or minima, and calculus is the mathematical tool used to do this. The …rst section of the book is devoted to the theory of optimization, and it begins with basic calculus. It moves beyond basic calculus in two ways, though. First, economic problems often have agents simultaneously choosing the values of more than one variable. For example, consumers choose commodity bundles, not the amount of a single commodity. To analyze problems with several choice variables, we need multivariate calculus. Second, as illustrated in Figure 1.1, the problem is not just a simple maximization problem. Instead, consumers maximize utility subject to a budget constraint. We must …gure out how to perform constrained optimization. Finding the market-clearing price is an equilibrium problem. An equilibrium is simply a state in which there is no pressure for anything to change, and the market-clearing price is the one at which suppliers have no incentive to raise or lower their prices and consumers have no incentive to raise or lower their o¤ers. Solutions to games are also based on the concept of equilibrium. Graphically, equilibrium analysis requires …nding the intersection of two curves, as in Figure 1.2. Mathematically, it involves the solution of CHAPTER 1. ECON AND MATH 5 several equations in several unknowns. The branch of mathematics used for this is linear (or matrix) algebra, and so we must learn to manipulate matrices and use them to solve systems of equations. Economic exercises often involve comparative statics analysis, which involves …nding how the optimum or equilibrium changes when one of the underlying parameters changes. For example, how does a consumer’s optimal bundle change when the underlying commodity prices change? How does a …rm’s optimal output change when an input or an output price changes? How does the market-clearing price change when an input price changes? All of these questions are answered using comparative statics analysis. Mathematically, comparative statics analysis involves multivariable calculus, often in combination with matrix algebra. This makes it sound hard. It isn’t really. But getting you to the point where you can perform comparative statics analysis means going through these two parts of mathematics. Comparative statics analysis is also at the heart of empirical work, that is, econometrics. A typical empirical project involves estimating an equation that relates a dependent variable to a set of independent variables. The estimated equation then tells how the dependent variable changes, on average, when one of the independent variables changes. So, for example, if one estimates a demand equation in which quantity demanded is the dependent variable and the good’s price, some substitute good prices, some complement good prices, and income are independent variables, the resulting equation tells how much quantity demanded changes when income rises, for example. But this is a comparative statics question. A good empirical project uses some math to derive the comparative statics results …rst, and then uses data to estimate the comparative statics results second. Consequently, econometrics and comparative statics analysis go hand-in-hand. Econometrics itself is the task of …tting the best line to a set of data points, as in Figure 1.3. There is some math behind that task. Much of it is linear algebra, because matrices turn out to provide an easy way to present the relevant equations. A little bit of the math is calculus, because "best" implies "optimal," and we use calculus to …nd optima. Econometrics also requires a knowledge of probability and statistics, which is the third branch of mathematics we will study. PART I OPTIMIZATION (multivariate calculus) CHAPTER 2 Single variable optimization One feature that separates economics from the other social sciences is the premise that individual actors, whether they are consumers, …rms, workers, or government agencies, act rationally to make themselves as well o¤ as possible. In other words, in economics everybody maximizes something. So, doing mathematical economics requires an ability to …nd maxima and minima of functions. This chapter takes a …rst step using the simplest possible case, the one in which the agent must choose the value of only a single variable. In later chapters we explore optimization problems in which the agent chooses the values of several variables simultaneously. Remember that one purpose of this course is to introduce you to the mathematical tools and techniques needed to do economics at the graduate level, and that the other is to teach you to frame economic questions, and their answers, mathematically. In light of the second goal, we will begin with a graphical analysis of optimization and then …nd the math that underlies the graph. Many of you have already taken calculus, and this chapter concerns singlevariable, di¤erential calculus. One di¤erence between teaching calculus in 7 CHAPTER 2. SINGLE VARIABLE OPTIMIZATION 8 $ π* π(q) q q* Figure 2.1: A pro…t function with a maximum an economics course and teaching it in a math course is that economists almost never use trigonometric functions. The economy has cycles, but none regular enough to model using sines and cosines. So, we will skip trigonometric functions. We will, however, need logarithms and exponential functions, and they are introduced in this chapter. 2.1 A graphical approach Consider the case of a competitive …rm choosing how much output to produce. When the …rm produces and sells q units it earns revenue R(q) and incurs costs of C(q). The pro…t function is (q) = R(q) C(q): The …rst term on the right-hand side is the …rm’s revenue, and the second term is its cost. Pro…t, as always, is revenue minus cost. More importantly for this chapter, Figure 2.1 shows the …rm’s pro…t function. The maximum level of pro…t is , which is achieved when output is q . Graphically this is very easy. The question is, how do we do it with equations instead? Two features of Figure 2.1 stand out. First, at the maximum the slope of the pro…t function is zero. Increasing q beyond q reduces pro…t, and CHAPTER 2. SINGLE VARIABLE OPTIMIZATION 9 $ π(q) q Figure 2.2: A pro…t function with a minimum decreasing q below q also reduces pro…t. Second, the pro…t function rises up to q and then falls. To see why this is important, compare it to Figure 2.2, where the pro…t function has a minimum. In Figure 2.2 the pro…t function falls to the minimum then rises, while in Figure 2.1 it rises to the maximum then falls. To make sure we have a maximum, we have to make sure that the pro…t function is rising then falling. This leaves us with several tasks. (1) We must …nd the slope of the pro…t function. (2) We must …nd q by …nding where the slope is zero. (3) We must make sure that pro…t really is maximized at q , and not minimized. (4) We must relate our …ndings back to economics. 2.2 Derivatives The derivative of a function provides its slope at a point. It can be denoted in two ways: f 0 (x) or df (x)=dx. The derivative of the function f at x is de…ned as df (x) f (x + h) f (x) = lim : (2.1) h!0 dx h The idea is as follows, with the help of Figure 2.3. Suppose we start at x and consider a change to x + h. Then f changes from f (x) to f (x + h). The ratio of the change in f to the change in x is a measure of the slope: 10 CHAPTER 2. SINGLE VARIABLE OPTIMIZATION f(x) f(x) f(x+h) f(x) x x x+h Figure 2.3: Approximating the slope of a function [f (x + h) f (x)]=[(x + h) x]. Make the change in x smaller and smaller to get a more precise measure of the slope, and, in the limit, you end up with the derivative. Finding the derivative comes from applying the formula in equation (2.1). And it helps to have a few simple rules in hand. We present these rules as a series of theorems. Theorem 1 Suppose f (x) = a. Then f 0 (x) = 0. Proof. f (x + h) h!0 h a a = lim h!0 h = 0: f 0 (x) = lim f (x) Graphically, a constant function, that is, one that yields the same value for every possible x, is just a horizontal line, and horizontal lines have slopes of zero. The theorem says that the derivative of a constant function is zero. Theorem 2 Suppose f (x) = x. Then f 0 (x) = 1. CHAPTER 2. SINGLE VARIABLE OPTIMIZATION 11 Proof. f (x + h) f (x) h!0 h (x + h) x = lim h!0 h h = lim h!0 h = 1: f 0 (x) = lim Graphically, the function f (x) = x is just a 45-degree line, and the slope of the 45-degree line is one. The theorem con…rms that the derivative of this function is one. Theorem 3 Suppose f (x) = au(x). Then f 0 (x) = au0 (x). Proof. f (x + h) f (x) h!0 h au(x + h) au(x) = lim h!0 h u(x + h) u(x) = a lim h!0 h 0 = au (x): f 0 (x) = lim This theorem provides a useful rule. When you multiply a function by a scalar (or constant), you also multiply the derivative by the same scalar. Graphically, multiplying by a scalar rotates the curve. Theorem 4 Suppose f (x) = u(x) + v(x). Then f 0 (x) = u0 (x) + v 0 (x). Proof. f (x + h) f (x) h!0 h [u(x + h) + v(x + h)] [u(x) + v(x)] = lim h!0 h u(x + h) u(x) v(x + h) u(x) = lim + h!0 h h 0 0 = u (x) + v (x): f 0 (x) = lim 12 CHAPTER 2. SINGLE VARIABLE OPTIMIZATION This rule says that the derivative of a sum is the sum of the derivatives. The next theorem is the product rule, which tells how to take the derivative of the product of two functions. Theorem 5 Suppose f (x) = u(x) v(x). Then f 0 (x) = u0 (x)v(x)+u(x)v 0 (x). Proof. f (x + h) f (x) h [u(x + h)v(x + h)] [u(x)v(x)] = lim h!0 h [u(x + h) u(x)]v(x) u(x + h)[v(x + h) = lim + h!0 h h f 0 (x) = lim h!0 v(x)] where the move from line 2 to line 3 entails adding then subtracting limh!0 u(x+ h)v(x)=h. Remembering that the limit of a product is the product of the limits, the above expression reduces to [v(x + h) [u(x + h) u(x)] v(x) + lim u(x + h) h!0 h!0 h h 0 0 = u (x)v(x) + u(x)v (x): f 0 (x) = lim v(x)] We need a rule for functions of the form f (x) = 1=u(x), and it is provided in the next theorem. Theorem 6 Suppose f (x) = 1=u(x). Then f 0 (x) = u0 (x)=[u(x)]2 . Proof. f (x + h) h!0 h f 0 (x) = lim = lim 1 u(x+h) f (x) 1 u(x) h u(x) u(x + h) = lim h!0 h[u(x + h)u(x)] [u(x + h) u(x)] 1 lim = lim h!0 u(x + h)u(x) h!0 h 1 = u0 (x) : [u(x)]2 h!0 CHAPTER 2. SINGLE VARIABLE OPTIMIZATION 13 Our …nal rule concerns composite functions, that is, functions of functions. This rule is called the chain rule. Theorem 7 Suppose f (x) = u(v(x)). Then f 0 (x) = u0 (v(x)) v 0 (x). Proof. First suppose that there is some sequence h1 ; h2 ; ::: with limi!1 hi = 0 and v(x + hi ) v(x) 6= 0 for all i. Then f (x + h) f (x) h!0 h u(v(x + h)) u(v(x)) lim h!0 h u(v(x + h)) u(v(x)) v(x + h) v(x) lim h!0 v(x + h) v(x) h v(x + h) v(x) u(v(x) + k)) u(v(x)) lim lim h!0 k!0 k h 0 0 u (v(x)) v (x): f 0 (x) = lim = = = = Now suppose that there is no sequence as de…ned above. Then there exists a sequence h1 ; h2 ; ::: with limi!1 hi = 0 and v(x + hi ) v(x) = 0 for all i. Let b = v(x) for all x,and f (x + h) f (x) h!0 h u(v(x + h)) u(v(x)) = lim h!0 h u(b) u(b) = lim h!0 h = 0: f 0 (x) = lim But u0 (v(x)) v 0 (x) = 0 since v 0 (x) = 0, and we are done. Combining these rules leads to the following really helpful rule: d a[f (x)]n = an[f (x)]n 1 f 0 (x): dx (2.2) 14 CHAPTER 2. SINGLE VARIABLE OPTIMIZATION This holds even if n is negative, and even if n is not an integer. So, for example, the derivative of xn is nxn 1 , and the derivative of (2x + 1)5 is 10(2x + 1)4 . The derivative of (4x2 1) :4 is :4(4x2 1) 1:4 (8x). Combining the rules also gives us the familiar division rule: u0 (x)v(x) v 0 (x)u(x) d u(x) : = dx v(x) [v(x)]2 (2.3) To get it, rewrite u(x)=v(x) as u(x) [v(x)] 1 . We can then use the product rule and expression (2.2) to get d u(x)v 1 (x) dx = u0 (x)v 1 (x) + ( 1)u(x)v 2 (x)v 0 (x) u0 (x) = v(x) v 0 (x)u(x) : v 2 (x) Multiplying both the numerator and denominator of the …rst term by v(x) yields (2.3). Getting more ridiculously complicated, consider f (x) = (x3 + 2x)(4x x3 1) : To di¤erentiate this thing, split f into three component functions, f1 (x) = x3 + 2x, f2 (x) = 4x 1, and f3 (x) = x3 . Then f (x) = f1 (x) f2 (x)=f3 (x), and f10 (x)f2 (x) f1 (x)f20 (x) f1 (x)f2 (x)f30 (x) 0 f (x) = + : f3 (x) f3 (x) [f3 (x)]2 We can di¤erentiate the component functions to get f1 (x) = 3x2 + 2, f20 (x) = 4, and f30 (x) = 3x2 . Plugging this all into the formula above gives us (3x2 + 2)(4x f (x) = x3 0 2.3 1) 4(x3 + 2x) + x3 3(x3 + 2x)(4x x6 1)x2 : Uses of derivatives In economics there are three major uses of derivatives. The …rst use comes from the economics idea of "marginal this" and "marginal that." In principles of economics courses, for example, marginal cost is CHAPTER 2. SINGLE VARIABLE OPTIMIZATION 15 de…ned as the additional cost a …rm incurs when it produces one more unit of output. If the cost function is C(q), where q is quantity, marginal cost is C(q + 1) C(q). We could divide output up into smaller units, though, by measuring in grams instead of kilograms, for example. Continually dividing output into smaller and smaller units of size h leads to the de…nition of marginal cost as c(q + h) c(q) M C(q) = lim : h!0 h Marginal cost is simply the derivative of the cost function. Similarly, marginal revenue is the derivative of the revenue function, and so on. The second use of derivatives comes from looking at their signs (the astrology of derivatives). Consider the function y = f (x). We might ask whether an increase in x leads to an increase in y or a decrease in y. The derivative f 0 (x) measures the change in y when x changes, and so if f 0 (x) 0 we know that y increases when x increases, and if f 0 (x) 0 we know that y decreases when x increases. So, for example, if the marginal cost function M C(q) or, equivalently, C 0 (q) is positive we know that an increase in output leads to an increase in cost. The third use of derivatives is for …nding maxima and minima of functions. This is where we started the chapter, with a competitive …rm choosing output to maximize pro…t. The pro…t function is (q) = R(q) C(q). As we saw in Figure 2.1, pro…t is maximized when the slope of the pro…t function is zero, or d = 0: dq This condition is called a …rst-order condition, often abbreviated as FOC. Using our rules for di¤erentiation, we can rewrite the FOC as d = R0 (q ) dq C 0 (q ) = 0; (2.4) which reduces to the familiar rule that a …rm maximizes pro…t by producing where marginal revenue equals marginal cost. Notice what we have done here. We have not used numbers or speci…c functions and, aside from homework exercises, we rarely will. Using general functions leads to expressions involving general functions, and we want to interpret these. We know that R0 (q) is marginal revenue and C 0 (q) is marginal cost. We end up in the same place we do using graphs, which is a good CHAPTER 2. SINGLE VARIABLE OPTIMIZATION 16 thing. The power of the mathematical approach is that it allows us to apply the same techniques in situations where graphs will not work. 2.4 Maximum or minimum? Figure 2.1 shows a pro…t function with a maximum, but Figure 2.2 shows one with a minimum. Both of them generate the same …rst-order condition: d =dq = 0. So what property of the function tells us that we are getting a maximum and not a minimum? In Figure 2.1 the slope of the curve decreases as q increases, while in Figure 2.2 the slope of the curve increases as q increases. Since slopes are just derivatives of the function, we can express these conditions mathematically by taking derivatives of derivatives, or second derivatives. The second derivative of the function f (x) is denoted f 00 (x) or d2 f =dx2 . For the function to have a maximum, like in Figure 2.1, the derivative should be decreasing, which means that the second derivative should be negative. For the function to have a minimum, like in Figure 2.2, the derivative should be increasing, which means that the second derivative should be positive. Each of these is called a second-order condition or SOC. The second-order condition for a maximum is f 00 (x) 0, and the second-order condition for a minimum is f 00 (x) 0. We can guarantee that pro…t is maximized, at least locally, if 00 (q ) 0. We can guarantee that pro…t is maximized globally if 00 (q) 0 for all possible values of q. Let’s look at the condition a little more closely. The …rst derivative of the pro…t function is 0 (q) = R0 (q) C 0 (q) and the second derivative is 00 (q) = R00 (q) C 00 (q). The second-order condition for a maximum is 00 (q) 0, which holds if R00 (q) 0 and C 00 (q) 0. So, we can guarantee that pro…t is maximized if the second derivative of the revenue function is nonpositive and the second derivative of the cost function is nonnegative. Remembering that C 0 (q) is marginal cost, the condition C 00 (q) 0 means that marginal cost is increasing, and this has an economic interpretation: each additional unit of output adds more to total cost than any unit preceding it. The condition R00 (q) 0 means that marginal revenue is decreasing, which means that the …rm earns less from each additional unit it sells. One special case that receives considerable attention in economics is the one in which R(q) = pq, where p is the price of the good. This is the CHAPTER 2. SINGLE VARIABLE OPTIMIZATION 17 revenue function for a price-taking …rm in a perfectly competitive industry. Then R0 (q) = p and R00 (q) = 0, and the …rst-order condition for pro…t maximization is p C 0 (q) = 0, which is the familiar condition that price equals marginal cost. The second-order condition reduces to C 00 (q) 0, which says that marginal cost must be nondecreasing. 2.5 Logarithms and the exponential function The functions ln x and ex turn out to play an important role in economics. The …rst is the natural logarithm, and the second is the exponential function. They are related: ln ex = eln x = x. The number e 2:718. Without going into why these functions are special for economics, let me show you why they are special for math. We know that d xn = xn 1 . dx n We can get the function x2 by di¤erentiating x3 =3, the function x by di¤erentiating x2 =2, the function x 2 by di¤erentiating x 1 , the function x 3 by di¤erentiating x 2 =2, and so on. But how can we get the function x 1 ? We cannot get it by di¤erentiating x0 =0, because that expression does not exist. We cannot get it by di¤erentiating x0 , because dx0 =dx = 0. So how do we get x 1 as a derivative? The answer is the natural logarithm: d 1 ln x = . dx x Logarithms have two additional useful properties: ln xy = ln x + ln y: and ln(xa ) = a ln x: Combining these yields ln(xa y b ) = a ln x + b ln y: (2.5) CHAPTER 2. SINGLE VARIABLE OPTIMIZATION 18 The left-hand side of this expression is non-linear, but the right-hand side is linear in the logarithms, which makes it easier to work with. Economists often use the form in (??) for utility functions and production functions. The exponential function ex also has an important di¤erentiation property: it is its own derivative, that is, d x e = ex : dx This implies that the derivative of eu(x) = u0 (x)eu(x) . 2.6 Problems 1. Compute the derivatives of the following functions: (a) f (x) = 12(x3 + 1)2 + 3 ln x2 (b) f (x) = 1=(4x (c) f (x) = e 5x 4 2)5 14x3 +2x (d) f (x) = (9 ln x)=x0:3 (e) f (x) = ax2 b cx d 2. Compute the derivative of the following functions: (a) f (x) = 12(x 1)2 (b) g(x) = (ln 3x)=(4x2 ) (c) h(x) = 1=(3x2 (d) f (x) = xe 2x + 1)4 x (e) g(x) = (2x2 p 3) 5x3 + 6 8 9x 3. Use the de…nition of the derivative (expression 2.1) to show that the derivative of x2 is 2x. 4. Use the de…nition of a derivative to prove that the derivative of 1=x is 1=x2 . 19 CHAPTER 2. SINGLE VARIABLE OPTIMIZATION 5. Answer the following: (a) Is f (x) = 2x3 12x2 increasing or decreasing at x = 3? (b) Is f (x) = ln x increasing or decreasing at x = 13? (c) Is f (x) = e x x1:5 increasing or decreasing at x = 4? (d) Is f (x) = 4x 1 x+2 increasing or decreasing at x = 2? 6. Answer the following: 2)=(4x + x2 ) increasing or decreasing at x = (a) Is f (x) = (3x 1? (b) Is f (x) = 1= ln x increasing or decreasing at x = e? (c) Is f (x) = 5x2 + 16x 12 increasing or decreasing at x = 6? 7. Optimize the following functions, and tell whether the optimum is a local maximum or a local minimum: (a) f (x) = 4x2 + 10x (b) f (x) = 120x0:7 (c) f (x) = 4x 6x 3 ln x 8. Optimize the following functions, and tell whether the optimum is a local maximum or a local minimum: (a) f (x) = 4x2 24x + 132 (b) f (x) = 20 ln x (c) f (x) = 36x 4x (x + 1)=(x + 2) 9. Consider the function f (x) = ax2 + bx + c. (a) Find conditions on a, b, and c that guarantee that f (x) has a unique global maximum. (b) Find conditions on a, b, and c that guarantee that f (x) has a unique global minimum. CHAPTER 2. SINGLE VARIABLE OPTIMIZATION 20 10. Beth has a minion (named Henry) and bene…ts when the minion exerts e¤ort, with minion e¤ort denoted by m. Her bene…t from m units of minion e¤ort is given by the function b(m). The minion does not like exerting e¤ort, and his cost of e¤ort is given by the function c(m). (a) Suppose that Beth is her own minion and that her e¤ort cost function is also c(m). Find the equation determining how much e¤ort she would exert, and interpret it. (b) What are the second-order condtions for the answer in (a) to be a maximum? (c) Suppose that Beth pays the minion w per unit of e¤ort. Find the equation determining how much e¤ort the minion will exert, and interpret it. (d) What are the second-order conditions for the answer in (c) to be a maximum? 11. A …rm (Bilco) can use its manufacturing facility to make either widgets or gookeys. Both require labor only. The production function for widgets is W = 20L1=2 and the production function for gookeys is G = 30L. The wage rate is $11 per unit of time, and the prices of widgets and gokeys are $9 and $3 per unit, repsectively. The manufacturing facility can accomodate 60 workers and no more. How much of each product should Bilco produce per unit of time? (Hint: If Bilco devotes L units of labor to widget production it has 60 L units of labor to devote to gookey production, and its pro…t function is (L) = 9 20L1=2 + 3 30(60 L) 11 60.) CHAPTER 3 Optimization with several variables Almost all of the intuition behind optimization comes from looking at problems with a single choice variable. In economics, though, problems often involve more than one choice variable. For example, consumers choose bundles of commodities, so must choose amounts of several di¤erent goods simultaneously. Firms use many inputs and must choose their amounts simultaneously. This chapter addresses issues that arise when there are several variables. The previous chapter used graphs to generate intuition. We cannot do that here because I am bad at drawing graphs with more than two dimensions. Instead, our intuition will come from what we learned in the last chapter. 3.1 A more complicated pro…t function In the preceding chapter we looked at a pro…t function in which the …rm chose how much output to produce. This time, instead of focusing on outputs, let’s focus on inputs. Suppose that the …rm can use n di¤erent inputs, and 21 CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 22 denote the amounts by x1 ; :::; xn . When the …rm uses x1 units of input 1, x2 units of input 2, and so on, its output is given by the production function Q = F (x1 ; :::; xn ): Inputs are costly, and we will assume that the …rm can purchase as much of input i as it wants for price ri , and it can sell as much of its output as it wants at the competitive price p. How much of each input should the …rm use to maximize pro…t? We know what to do when there is only one input (n = 1). Call the input labor (L) and its price the wage (w). The production function is then Q = F (L). When the …rm employs L units of labor it produces F (L) units of output and sells them for p units each, for revenue of pF (L). Its only cost is a labor cost equal to wL because it pays each unit of labor the wage w. Pro…t, then, is (L) = pF (L) wL. The …rst-order condition is 0 (L) = pF 0 (L) w = 0; which can be interpreted as the …rm’s pro…t maximizing labor demand equating the value marginal product of labor pF 0 (L) to the wage rate. Using one additional unit of labor costs an additional w but increases output by F 0 (L), which increases revenue by pF 0 (L). The …rm employs labor as long as each additional unit generates more revenue than it costs, and stops when the added revenue and the added cost exactly o¤set each other. What happens if there are two inputs (n = 2), call them capital (K) and labor (L)? The production function is then Q = F (K; L), and the corresponding pro…t function is (K; L) = pF (K; L) rK wL: (3.1) How do we …nd the …rst-order condition? That is the task for this chapter. 3.2 Vectors and Euclidean space Before we can …nd a …rst-order condition for (3.1), we …rst need some terminology. A vector is an array of n numbers, written (x1 ; :::; xn ). In our example of the input-choosing pro…t-maximizing …rm, the vector (x1 ; :::; xn ) is an input vector. For each i between 1 and n, the quantity xi is the amount of the i-th input. More generally, we call xi the i-th component of CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 23 the vector (x1 ; :::; xn ). The number of components in a vector is called the dimension of the vector; the vector (x1 ; :::; xn ) is n-dimensional. Vectors are collections of numbers. They are also numbers themselves, and it will help you if you begin to think of them this way. The set of real numbers is commonly denoted by R, and we depict R using a number line. We can depict a 2-dimensional vector using a coordinate plane anchored by two real lines. So, the vector (x1 ; x2 ) is in R2 , which can be thought of as R R. We call R2 the 2-dimensional Euclidean space. When you took plane geometry in high school, this was Euclidean geometry. When a vector has n components, it is in Rn , or n-dimensional Euclidean space. In this text vectors are sometimes written out as (x1 ; :::; xn ), but sometimes that is cumbersome. We use the symbol x to denote the vector whose components are x1 ; :::; xn . That way we can talk about operations involving two vectors, like x and y. Three common operations are used with vectors. We begin with addition: x + y = (x1 + y1 ; x2 + y2 ; :::; xn + yn ): Adding vectors is done component-by-component. Multiplication is more complicated, and there are two notions. One is scalar multiplication. If x is a vector and a is a scalar (a real number), then ax = (ax1 ; ax2 ; :::; axn ). Scalar multiplication is achieved by multiplying each component of the vector by the same number, thereby either "scaling up" or "scaling down" the vector. Vector subtraction can be achieved through addition and using 1 as the scalar: x y = x + ( 1)y. The other form of multiplication is the inner product, sometimes called the dot product. It is done using the formula x y = x1 y1 + x2 y2 + ::: + xn yn . Vector addition takes two vectors and yields another vector, and scalar multiplication takes a vector and a scalar and yields another vector. But the inner product takes two vectors and yields a scalar, or real number. You might wonder why we would ever want such a thing. Here is an example. Suppose that a …rm uses n inputs in amounts x1 ; :::; xn . It pays ri per unit of input i. What is its total production cost? Obviously, it is r1 x1 + ::: + rn xn , which can be easily written as r x. CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 24 Similarly, if a consumer purchases a commodity bundle given by the vector x = (x1 ; :::; xn ) and pays prices given by the vector p = (p1 ; :::; pn ), her total expenditure is p x. Often it is more convenient to leave the "dot" out of the inner product, and just p write px. A second use comes from looking at 2 2 x x = x1 + ::: + xn . Then x x is the distance from the point x (remember, it’s a number) to the origin. This is also called the norm of the vector x, 1 and it is written kxk = (x x) 2 . Both vector addition and the inner product are commutative, that is, they do not depend on the order in which the two vectors occur. This will contrast with matrices in a later chapter, where matrix multiplication is dependent on the order in which the matrices are written. Vector analysis also requires some de…nitions for ordering vectors. For real numbers we have the familiar relations >, , =, , and <. For vectors, and x = y if xi = yi for all i = 1; ::; n; x y if xi yi for all i = 1; :::; n; x > y if x y but x 6= y; x y if xi > yi for all i = 1; :::; n. From the third one it follows that x > y if xi yi for all i = 1; :::; n and xi > yi for some i between 1 and n. The fourth condition can be read x is strictly greater than y component-wise. 3.3 Partial derivatives The trick to maximizing a function of several variables, like (3.1), is to maximize it according to each variable separately, that is, by …nding a …rst-order condition for the choice of K and another one for the choice of L. In general both of these conditions will depend on the values of both K and L, so we will have to solve some simultaneous equations. We will get to that later. The point is that we want to di¤erentiate (3.1) once with respect to K and once with respect to L. Di¤erentiating a function of two or more variables with respect to only one of them is called partial di¤erentiation. Let f (x1 ; :::; xn ) be a general function of n variables. The i-th partial derivative of f is @f f (x1 ; x2 ; :::; xi 1 ; xi + h; xi+1 ; :::; xn ) f (x1 ; :::; xn ) (x) = lim : (3.2) h!0 @xi h CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 25 This de…nition might be a little easier to see with one more piece of notation. The coordinate vector ei is the vector with components given by eii = 1 and eij = 0 when j 6= i. The …rst coordinate vector is e1 = (1; 0; :::; 0), the second coordinate vector is e2 = (0; 1; 0; :::; 0), and so on through the n-th coordinate vector en = (0; :::; 0; 1). So, coordinate vector ei has a one in the i-th place and zeros everywhere else. Using coordinate vectors, the de…nition of the i-th partial derivative in (3.2) can be rewritten @f f (x + hei ) (x) = lim h!0 @xi h f (x) : The i-th partial derivative of the function f is simply the derivative one gets by holding all of the components …xed except for the i-th component. One takes the partial by pretending that all of the other variables are really just constants and di¤erentiating as if it were a single-variable function. For example, consider the function f (x1 ; x2 ) = (5x1 2)(7x2 3)2 . The partial derivatives are f1 (x1 ; x2 ) = 5(7x2 3)2 and f2 (x1 ; x2 ) = 14(5x1 2)(7x2 3). We sometimes use the notation fi (x) to denote @f (x)=@xi . When a function is de…ned over n-dimensional vectors it has n di¤erent partial derivatives. It is also possible to take partial derivatives of partial derivatives, much like second derivatives in single-variable calculus. We use the notation fij (x) = @2f (x): @xi @xj We call fii (x) the second partial of f with respect to xi , and we call fij (x) the cross partial of f (x) with respect to xi and xj . It is important to note that, in most cases, fij (x) = fji (x); that is, the order of di¤erentiation does not matter. In fact, this result is important enough to have a name: Young’s Theorem. Partial di¤erentiation requires a restatement of the chain rule: Theorem 8 Consider the function f : Rn ! R given by f (u1 (x); u2 (x); :::; un (x)); where u1 ; :::; un are functions of the one-dimensional variable x. Then df = f1 u01 (x) + f2 u02 (x) + ::: + fn u0n (x) dx 26 CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES This rule is best explained using an example. Suppose that the function is f (3x2 2; 5 ln x). It’s derivative with respect to x is d f (3x2 dx 2; 5 ln x) = f1 (3x2 2; 5 ln x) (6x) + f2 (3x2 2; 5 ln x) 5 x : The basic rule to remember is that when variable we are di¤erentiating with respect to appears in several places in the function, we di¤erentiate with respect to each argument separately and then add them together. The following lame example shows that this works. Let f (y1 ; y2 ) = y1 y2 , but the values y1 and y2 are both determined by the value of x, with y1 = 2x and y2 = 3x2 . Substituting we have f (x) = (2x)(3x2 ) = 6x3 df = 18x2 : dx But, if we use the chain rule, we get dy1 dy2 df = y2 + y1 dx dx dx = (2)(3x2 ) + (6x)(2x) = 18x2 : It works. 3.4 Multidimensional optimization Let’s return to our original problem, maximizing the pro…t function given in expression (3.1). The …rm chooses both capital K and labor L to maximize (K; L) = pF (K; L) rK wL: Think about the …rm as solving two problems simultaneously: (i) given the optimal amount of labor, L , the …rm wants to use the amount of capital that maximizes (K; L ); and (ii) given the optimal amount of capital, K , the …rm wants to employ the amount of labor that maximizes (K ; L). Problem (i) translates into @ (K ; L ) = 0 @K CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 27 and problem (ii) translates into @ (K ; L ) = 0: @L Thus, optimization in several dimensions is just like optimization in each single dimension separately, with the provision that all of the optimization problems must be solved together. The two equations above are the …rstorder conditions for the pro…t-maximization problem. To see how this works, suppose that the production function is F (K; L) = 1=2 K + L1=2 , that the price of the good is p = 10, the price of capital is r = 5, and the wage rate is w = 4. Then the pro…t function is (K; L) = 10(K 1=2 + L1=2 ) 5K 4L. The …rst-order conditions are @ (K; L) = 5K @K 1=2 5=0 and @ (K; L) = 5L 1=2 4 = 0: @L The …rst equation gives us K = 1 and the second gives us L = 25=16. In this example the two …rst-order conditions were independent, that is, the FOC for K did not depend on L and the FOC for L did not depend on K. This is not always the case, as shown by the next example. Example 1 The production function is F (K; L) = K 1=4 L1=2 , the price is 12, the price of capital is r = 6, and the wage rate is w = 6. Find the optimal values of K and L. Solution. The pro…t function is (K; L) = 12K 1=4 L1=2 6K 6L: The …rst-order conditions are @ (K; L) = 3K @K and 3=4 @ (K; L) = 6K 1=4 L @L L1=2 6=0 (3.3) 1=2 6 = 0: (3.4) CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 28 To solve these, note that (3.4) can be rearranged to get K 1=4 = 1 L1=2 K 1=4 = L1=2 K = L2 where the last line comes from raising both sides to the fourth power. Plugging this into (3.3) yields L1=2 K 3=4 L1=2 (L2 )3=4 L1=2 L3=2 1 L = 2 = 2 = 2 = 2 1 : 2 Plugging this back into K = L2 yields K = 1=4. L = This example shows the steps for solving a multi-dimensional optimization problem. Now let’s return to the general problem to see what the …rst-order conditions tell us. The general pro…t-maximization problem is max pF (x1 ; :::; xn ) x1 ;:::;xn r1 x 1 ::: rn x n or, in vector notation, max pF (x) x r x: The …rst-order conditions are: pF1 (x1 ; :::; xn ) .. . r1 = 0 pFn (x1 ; :::; xn ) rn = 0. The i-th FOC is pFi (x) = ri , which is the condition that the value marginal product of input i equals its price. This is the same as the condition for a single variable, and it holds for every input. 29 CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 3.5 Comparative statics analysis Being able to do multivariate calculus allows us to do one of the most important tasks in microeconomics: comparative statics analysis. The standard comparative statics questions is, "How does the optimum change when one of the underlying variables changes?" For example, how does the …rm’s demand for labor change when the output price changes, or when the wage rate changes? This is an important problem, and many papers (and dissertations) have relied on not much more than comparative statics analysis. If there is one tool you have in your kit at the end of the course, it should be comparative statics analysis. To see how it works, let’s return to the pro…t maximization problem with a single choice variable: max pF (L) wL: L The FOC is pF 0 (L) w = 0: (3.5) The comparative statics question is, how does the optimal value of L change when p changes? To answer this, let’s …rst assume that the marginal product of labor is strictly decreasing, so that F 00 (L) < 0: Note that this guarantees that the second-order condition for a maximization is satis…ed. The trick we now take is to implicitly di¤erentiate equation (3.5) with respect to p, treating L as a function of p. In other words, rewrite the FOC so that L is replaced by the function L (p): pF 0 (L (p)) w=0 and di¤erentiate both sides of the expression with respect to p. We get F 0 (L (p)) + pF 00 (L (p)) dL = 0: dp The comparative statics question is now simply the astrology question, "What is the sign of dL =dp?" Rearranging the above equation to isolate dL =dp on the left-hand side gives us dL = dp F 0 (L ) : pF 00 (L ) CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 30 We know that F 0 (L) > 0 because production functions are increasing, and we know that pF 00 (L ) < 0 because we assumed strictly diminishing marginal product of labor, i.e. F 00 (L) < 0. So, dL =dp has the form of the negative of a ratio of a positive number to a negative number, which is positive. This tells us that the …rm demands more labor when the output price rises, which makes sense: when the output price rises producing output becomes more pro…table, and so the …rm wants to expand its operation to generate more pro…t. We can write the comparative statics problem generally. Suppose that the objective function, that is, the function the agent wants to optimize, is f (x; s), where x is the choice variable and s is a shift parameter. Assume that the second-order condition holds strictly, so that fxx (x; s) < 0 for a maximization problem and fxx (x; s) > 0 for a minimization problem. These conditions guarantee that there is no "‡at spot" in the objective function, so that there is a unique solution to the …rst-order condition. Let x denote the optimal value of x. The comparative statics question is, "What is the sign of dx =ds?" To get the answer, …rst derive the …rst-order condition: fx (x; s) = 0: Next implicitly di¤erentiate with respect to s to get fxx (x; s) dx + fxs (x; s) = 0: ds Rearranging yields the comparative statics derivative dx = ds fxs (x; s) : fxx (x; s) (3.6) We want to know the sign of this derivative. For a maximization problem we have fxx < 0 by assumption and so the negative sign cancels out the sign of the denominator. Consequently, the sign of the comparative statics derivative dx =ds is the same as the sign of the numerator, fxs (x; s). For a minimization problem we have fxx > 0, and so the comparative statics derivative dx =ds has the opposite sign from the partial derivative fxs (x; s). Example 2 A person must decide how much to work in a single 24-hour day. She gets paid w per hour of labor, but gets utility from each hour she CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 31 does not spend at work. This utility is given by the function u(t), where t is the amount of time spent away from work, and u has the properties u0 (t) > 0 and u00 (t) < 0. Does the person work more or less when her wage increases? Solution. Let L denote the amount she works, so that 24 L is is the amount of time she does not spend at work. Her utility from this leisure time is therefore u(24 L), and her objective is maxwL + u(24 L L): The FOC is u0 (24 w L ) = 0: Write L as a function of w to get L (w) and rewrite the FOC as: w u0 (24 L (w)) = 0: Di¤erentiate both sides with respect to w (this is implicit di¤erentiation) to get dL 1 + u00 (24 L ) = 0: dw Solving for the comparative statics derivative dL =dw yields dL = dw 1 u00 (24 L) > 0: She works more when the wage rate w increases. 3.5.1 An alternative approach (that I don’t like) Many people use an alternative approach to comparative statics analysis. It gets to the same answers, but I do not like this approach as much. We will get to why later. The approach begins with total di¤erential, and the total di¤erential of the function g(x1 ; :::; xn ) is dg = g1 dx1 + ::: + gn dxn : We want to use total di¤erentials to get comparative statics derivatives. CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 32 Remember our comparative statics problem: we choose x to optimize f (x; s). The FOC is fx (x; s) = 0: Let’s take the total di¤erential of both sides. The total di¤erential of the right-hand side is zero, and the total di¤erential of the left-hand side is d[fx (x; s)] = fxx dx + fxs ds: Setting the above expression equal to zero yields fxx dx + fxs ds = 0: The derivative we want is the comparative statics derivative dx=ds. We can solve for this expression in the above equation: fxx dx + fxs ds = 0 fxx dx = fxs ds fxs dx = : ds fxx (3.7) This is exactly the comparative statics derivative we found above in equation (3.6). So the method works, and many students …nd it straightforward and easier to use than implicit di¤erentiation. Let’s stretch our techniques a little and have a problem with two shift parameters, s and r, instead of just one. The problem is to optimize f (x; r; s), and the FOC is fx (x; r; s) = 0: If we want to do comparative statics analysis using our (preferred) implicit di¤erentiation approach, we would …rst write x as a function of the two shift parameters, so that fx (x(r; s); r; s) = 0: To …nd the comparative statics derivative dx=dr, we implicitly di¤erentiate with respect to r to get fxx dx + fxr = 0 dr dx fxr = : dr fxx CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 33 This should not be a surprise, since it is just like expression (3.6) except it replaces s with r. Using total di¤erentials, we would …rst take the total di¤erential of fx (x; r; s) to get fxx dx + fxr dr + fxs ds = 0: We want to solve for dx=dr, and doing so yields fxx dx + fxr dr + fxs ds = 0 fxx dx = fxr dr fxs ds dx fxr fxs ds = : dr fxx fxx dr On the face of it, this does not look like the same answer. But, both s and r are shift parameters, so s is not a function of r. That means that ds=dr = 0. Substituting this in yields fxr dx = dr fxx as expected. So what is the di¤erence between the two approaches? In the implicit di¤erentiation approach we recognized that s does not depend on r at the beginning of the process, and in the total di¤erential approach we recognized it at the end. So both work, it’s just a matter of when you want to do your remembering. All of that said, I still like the implicit di¤erentiation approach better. To see why, think about what the derivative dx=ds means. As we constructed it back in equation (2.1), dx=ds is the limit of x= s as s ! 0. According to this intuition, ds is the limit of s as it goes to zero, so ds is zero. But we divided by it in equation (3.7), and you were taught very young that you cannot divide by zero. So, on a purely mathematical basis, I object to the total di¤erential approach because it entails dividing by zero, and I prefer to think of dx=ds as a single entity with a long name, and not a ratio of dx and ds. On a practical level, though, the total di¤erential approach works just …ne. It’s just not going to show up anywhere else in this book. 3.6 Problems 1. Consider the vectors x = (4; 3; 6; 2) and y = (6; 1; 7; 7). CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES 34 (a) Write down the vector 2y + 3x: (b) Which of the following, if any, are true: x = y, x x y? y, x < y, or (c) Find the inner product x y. p p p (d) Is x x + y y (x + y) (x + y)? 2. Consider the vectors x = (5; 0; 6; 2) and y = (3; 2; 3; 2). (a) Write down the vector 6x 4y. (b) Find the inner product x y. p p p (c) Verify that x x + y y > (x + y) (x + y). 3. Consider the function f (x; y) = 4x2 + 3y 2 12xy + 18x. (a) Find the partial derivative fx (x; y). (b) Find the partial derivative fy (x; y). (c) Find the critical point of f . 4. Consider the function f (x; y) = 16xy 4x + 2=y. (a) Find the partial derivative fx (x; y). (b) Find the partial derivative fy (x; y). (c) Find the critical point of f . 5. Consider the function u(x; y) = 3 ln x + 2 ln y. (a) Write the equation for the indi¤erence curve corresponding to the utility level k. (b) Find the slope of the indi¤erence curve at point (x; y). 6. A …rm faces inverse demand function p(q) = 120 …rm’s output. Its cost function is cq. 4q, where q is the (a) Write down the …rm’s pro…t function. (b) Find the pro…t-maximizing level of pro…t as a function of the unit cost c. CHAPTER 3. OPTIMIZATION WITH SEVERAL VARIABLES (c) Find the comparative statics derivative dq=dc. negative? 35 Is it positive or (d) Write the maximized pro…t function as a function of c. (e) Find the derivative showing how pro…t changes when c changes. (f) Show that d =dc = q. 7. Find dx=da from each of the following expressions. (a) 15x2 + 3xa 5 x = 20: a (b) 6x2 a = 5a 5xa2 8. Each worker at a …rm can produce 4 units per hour, each worker must p be paid $w per hour, and the …rm’s revenue function is R(L) = 30 L, where L is the number of workers employed p (fractional workers are okay). The …rm’s pro…t function is (L) = 30 4L wL. (a) Show that L = 900=w2 : (b) Find dL =dw. What’s its sign? (c) Find d =dw. What’s its sign? 9. An isoquant is a curve showing the combinations of inputs that all lead to the same level of output. When the production function over capital K and labor L is F (K; L), an isoquant corresponding to 100 units of output is given by the equation F (K; L) = 100. (a) If capital is on the vertical axis and labor on the horizontal, …nd the slope of the isoquant. (b) Suppose that production is increasing in both capital and labor. Does the isoquant slope upward or downward? CHAPTER 4 Constrained optimization Microeconomics courses typically begin with either consumer theory or producer theory. If they begin with consumer theory the …rst problem they face is the consumer’s constrained optimization problem: the consumer chooses a commodity bundle, or vector of goods, to maximize utility without spending more than her budget. If the course begins with producer theory, the …rst problem it poses is the …rm’s cost minimization problem: the …rm chooses a vector of inputs to minimize the cost of producing a predetermined level of output. Both of these problems lead to a graph of the form in Figure 1.1. So far we have only looked at unconstrained optimization problems. But many problems in economics have constraints. The consumer’s budget constraint is a classic example. Without a budget constraint, and under the assumption that more is better, a consumer would choose an in…nite amount of every good. This obviously does not help us describe the real world, because consumers cannot purchase unlimited quantities of every good. The budget constraint is an extra condition that the optimum must satisfy. How do we make sure that we get the solution that maximizes utility while still letting the budget constraint hold? 36 37 CHAPTER 4. CONSTRAINED OPTIMIZATION x2 E x1 Figure 4.1: Consumer’s problem 4.1 A graphical approach Let’s look more carefully at the consumer’s two-dimensional maximization problem. The problem can be written as follows: maxu(x1 ; x2 ) x1 ;x2 s.t. p1 x1 + p2 x2 = M where pi is the price of good i and M is the total amount the consumer has to spend on consumption. The second line is the budget constraint, it says that total expenditure on the two goods is equal to M . The abbreviation "s.t." stands for "subject to," so that the problem for the consumer is to choose a commodity bundle to maximize utility subject to a budget constraint. Figure 4.1 shows the problem graphically, and it should be familiar. What we want to do in this section is …gure out what equations we need to characterize the solution. The optimal consumption point, E, is a tangency point, and it has two features: (i) it is where the indi¤erence curve is tangent to the budget line, and (ii) it is on the budget line. Let’s translate these into math. To …nd the tangency condition we must …gure out how to …nd the slopes of the two curves. We can do this easily using implicit di¤erentiation. Begin with the budget line, because it’s easier. Since x2 is on the vertical axis, we CHAPTER 4. CONSTRAINED OPTIMIZATION 38 want to …nd a slope of the form dx2 =dx1 . Treating x2 as a function of x1 and rewriting the budget constraint yields p1 x1 + p2 x2 (x1 ) = M: Implicit di¤erentiation gives us p1 + p2 dx2 =0 dx1 because the derivative of M with respect to x1 is zero. Rearranging yields dx2 = dx1 p1 : p2 (4.1) Of course, we could have gotten to the same place by rewriting the equation for the budget line in slope-intercept form, x2 = M=p2 (p1 =p2 )x1 , but we have to use implicit di¤erentiation anyway to …nd the slope of the indi¤erence curve, and it is better to apply it …rst to the easier case. Now let’s …nd the slope of the indi¤erence curve. The equation for an indi¤erence curve is u(x1 ; x2 ) = k for some scalar k. Treat x2 as a function of x1 and rewrite to get u(x1 ; x2 (x1 )) = k: Now implicitly di¤erentiate with respect to x1 to get @u(x1 ; x2 ) @u(x1 ; x2 ) dx2 + = 0: @x1 @x2 dx1 Rearranging yields dx2 = dx1 @u(x1 ;x2 ) @x1 @u(x1 ;x2 ) @x2 : (4.2) The numerator is the marginal utility of good 1, and the denominator is the marginal utility of good 2, so the slope of the indi¤erence curve is the negative of the ratio of marginal utilities, which is also known as the marginal rate of substitution. 39 CHAPTER 4. CONSTRAINED OPTIMIZATION Condition (i), that the indi¤erence curve and budget line are tangent, requires that the slope of the budget line in (4.1) is the same as the slope of the indi¤erence curve in (4.2), or u1 (x1 ; x2 ) = u2 (x1 ; x2 ) p1 : p2 (4.3) The other condition, condition (ii), says that the bundle (x1 ; x2 ) must lie on the budget line, which is simply p1 x1 + p2 x2 = M . (4.4) Equations (4.3) and (4.4) constitute two equations in two unknowns (x1 and x2 ), and so they completely characterize the solution to the consumer’s optimization problem. The task now is to characterize the solution in a more general setting with more dimensions. 4.2 Lagrangians The way that we solve constrained optimization problems is by using a trick developed by the 18-th century Italian-French mathematician Joseph-Louis Lagrange. (There is also a 1972 ZZ Top song called La Grange, so don’t get confused.) Suppose that our objective is to solve an n-dimensional constrained utility maximization problem: max u(x1 ; :::; xn ) x1 ;:::;xn s.t. p1 x1 + ::: + pn xn = M . Our …rst step is to set up the Lagrangian L(x1 ; :::; xn ; ) = u(x1 ; :::; xn ) + (M p1 x1 ::: pn xn ): This requires some interpretation. First of all, the variable is called the Lagrange multiplier (and the Greek letter is lambda). Second, let’s think about the quantity M p1 x1 ::: pn xn . It has to be zero according to the budget constraint, but suppose it was positive. What would it mean? M is income, and p1 x1 + ::: + pn xn is expenditure on consumption. Income minus expenditure is simply unspent income. But unspent income is measured in dollars, and utility is measured in utility units (or utils), so we cannot simply 40 CHAPTER 4. CONSTRAINED OPTIMIZATION add these together. The Lagrange multiplier converts the dollars into utils, and is therefore measured in utils/dollar. The expression (M p1 x1 ::: pn xn ) can be interpreted as the utility of unspent income. The Lagrangian, then, is the utility of consumption plus the utility of unspent income. The budget constraint, though, guarantees that there is no unspent income, and so the second term in the Lagrangian is necessarily zero. We still want it there, though, because it is important for …nding the right set of …rst-order conditions. Note that the Lagrangian has not only the xi ’s as arguments, but also the Lagrange multiplier . The …rst-order conditions arise from taking n + 1 partial derivatives of L, one for each of the xi ’s and one for : @L @u = p1 = 0 @x1 @x1 .. . @u @L = pn = 0 @xn @xn @L = M p1 x1 ::: @ (4.5a) (4.5b) pn xn = 0 (4.5c) Notice that the last FOC is simply the budget constraint. So, optimization using the Lagrangian guarantees that the budget constraint is satis…ed. Also, optimization using the Lagrangian turns the n-dimensional constrained optimization problem into an (n+1)-dimensional unconstrained optimization problem. These two features give the Lagrangian approach its appeal. 4.3 A 2-dimensional example The utility function is u(x1 ; x2 ) = x10:5 x20:5 , the prices are p1 = 10 and p2 = 20, and the budget is M = 120. The consumer’s problem is then 1=2 1=2 maxx1 x2 x1 ;x2 s.t. 10x1 + 20x2 = 120: What are the utility maximizing levels of x1 and x2 ? To answer this, we begin by setting up the Lagrangian 1=2 1=2 L(x1 ; x2 ; ) = x1 x2 + (120 10x1 20x2 ): CHAPTER 4. CONSTRAINED OPTIMIZATION 41 The …rst-order conditions are @L 1=2 1=2 = 0:5x1 x2 10 = 0 @x1 @L 1=2 1=2 = 0:5x1 x2 20 = 0 @x2 @L = 120 10x1 20x2 = 0 @ Of course, equation (4.6c) is simply the budget constraint. We have three equations in three unknowns (x1 , x2 , and ). them, …rst rearrange (4.6a) and (4.6b) to get 1 = 20 x2 x1 1 40 x1 x2 (4.6b) (4.6c) To solve 1=2 and = (4.6a) 1=2 : Set these equal to each other to get 1 20 x2 x1 1=2 1=2 2 x2 x1 1 = 40 x1 x2 = x1 x2 1=2 1=2 2x2 = x1 where the last line comes from cross-multiplying. Substitute x1 = 2x2 into (4.6c) to get 120 10(2x2 ) 20x2 = 0 40x2 = 120 x2 = 3: Because x1 = 2x2 , we have x1 = 6: CHAPTER 4. CONSTRAINED OPTIMIZATION 42 Finally, we know from the rearrangement of (4.6a) that 1 = 20 x2 x1 1 3 = 20 6 1 p : = 20 2 4.4 1=2 1=2 Interpreting the Lagrange multiplier Remember that we said that the second term in the Lagrangian is the utility value of unspent income, which, of course, is zero because there is no unspent income. This term is (M p1 x1 p2 x2 ). So, the Lagrange multiplier should be the marginal utility of (unspent) income, because it is the slope of the utility-of-unspent-income function. Let’s see if this is true. To do so, let’s generalize the problem so that income is M instead of 120. All of the steps are the same as above, so we still have x1 = 2x2 . Substituting into the budget constraint gives us M 10(2x2 ) 20x2 = 0 M x2 = 40 M x1 = 20 1 p : = 20 2 Plugging these numbers back into the utility function gives us u(x1 ; x2 ) = M 20 0:5 M 40 0:5 = M p : 20 2 Di¤erentiating this expression with respect to income M yields du 1 = p = ; dM 20 2 and the Lagrange multiplier really does measure the marginal utility of income. 43 CHAPTER 4. CONSTRAINED OPTIMIZATION In general, the Lagrange multiplier measures the marginal value of relaxing the constraint, where the units used to measure value are determined by the objective function. In our case the objective function is a utility function, so the marginal value is marginal utility. The constraint is relaxed by allowing income to be higher, so the Lagrange multiplier measures the marginal utility of income. Now think instead about a …rm’s cost-minimization problem. Let xi be the amount of input i employed by the …rm, let ri be its price, let F (x1 ; :::; xn ) be the production function, and let q be the desired level of output. The …rm’s problem would be min r1 x1 + ::: + rn xn x1 ;:::;xn s.t. F (x1 ; :::; xn ) = q The Lagrangian is then L(x1 ; :::; xn ; ) = r1 x1 + ::: + rn xn + (q F (x1 ; :::; xn )): Since the …rm is minimizing cost, reducing cost from the optimum would require reducing the output requirement q. So, relaxing the constraint is lowering q. The interpretation of is the marginal cost of output, which was referred to simply as marginal cost way back in Chapter 2. So, using the Lagrangian to solve the …rm’s cost minimization problem gives you the …rm’s marginal output cost function for free. 4.5 A useful example - Cobb-Douglas Economists often rely on the Cobb-Douglas class of functions which take the form f (x1 ; :::; xn ) = xa11 xa22 xann where all of the ai ’s are positive. The functional form arose out of a 1928 collaboration between economist Paul Douglas and mathematician Charles Cobb, and was designed to …t Douglas’s production data. To see its usefulness, consider a consumer choice problem with a Cobbxann : Douglas utility function u(x) = xa11 xa22 max u(x) x1 ;:::;xn 44 CHAPTER 4. CONSTRAINED OPTIMIZATION s.t. p1 x1 + ::: + pn xn = M: Form the Lagrangian L(x1 ; :::; xn ; ) = xa11 xann + (M p1 x1 The …rst-order conditions take the form @L ai = xa11 xann @xi xi ::: pn xn ): pi = 0 for i = 1; :::; n and @L =M @ p1 x1 ::: pn xn = 0; which is just the budget constraint. The expression for @L=@xi can be rearranged to become ai ai a 1 pi = u(x) x1 xann pi = 0: xi xi This yields that ai u(x) xi for i = 1; :::; n. Substitute these into the budget constraint: pi = M M p1 x1 ::: pn xn = 0 an u(x) a1 u(x) x1 ::: xn = 0 x1 xn u(x) (a1 + ::: + an ) = 0: M Now solve this for the Lagrange multiplier : = u(x) a1 + ::: + an : M Finally, plug this back into (4.7) to get ai u(x) 1 xi ai u(x) M = xi u(x)(a1 + ::: + an ) ai M = : a1 + ::: + an xi pi = (4.7) 45 CHAPTER 4. CONSTRAINED OPTIMIZATION Finally, solve this for xi to get the demand function for good i: xi = ai M : a1 + ::: + an pi (4.8) That was a lot of steps, but rearranging (4.8) yields an intuitive and easily memorizable expression. In fact, most graduate students in economics have memorized it by the end of their …rst semester because it turns out to be so handy. Rearrange (4.8) to pi xi ai = : M a1 + ::: + an The numerator of the left-hand side is the amount spent on good i. The denominator of the left-hand side is the total amount spent. The left-hand side, then, is the share of income spent on good i. The equation says that the share of spending is determined entirely by the exponents of the CobbDouglas utility function. In especially convenient cases the exponents sum to one, in which case the spending share for good i is just equal to the exponent on good i. The demand function in (4.8) lends itself to some particularly easy comparative statics analysis. The obvious comparative statics derivative for a demand function is with respect to its own price: dxi = dpi ai M a1 + ::: + an p2i 0 and so demand is downward-sloping, as it should be. Another comparative statics derivative is with respect to income: ai 1 dxi = dM a1 + ::: + an pi 0: All goods are normal goods when the utility function takes the Cobb-Douglas form. Finally, one often looks for the e¤ects of changes in the prices of other goods. We can do this by taking the comparative statics derivative of xi with respect to price pj , where j 6= i. dxi = 0: dpj This result holds because the other prices appear nowhere in the demand function (4.8), which is another feature that makes Cobb-Douglas special. 46 CHAPTER 4. CONSTRAINED OPTIMIZATION We can also use Cobb-Douglas functions in a production setting. Consider the …rm’s cost-minimization problem when the production function is Cobb-Douglas, so that F (x) = xa11 xann . This time, though, we are going to assume that a1 + ::: + an = 1. The problem is min p1 x1 + ::: + pn xn x1 ;:::;xn s.t. xa11 xann = q. Set up the Lagrangian xa11 L(x1 ; :::; xn ; ) = p1 x1 + ::: + pn xn + (q xann ): The …rst-order conditions take the form @L = pi @xi ai F (x) =0 xi for i = 1; :::; n and @L = q xa11 xann = 0; @ which is just the production constraint. Rearranging the expression for @L=@xi yields q x i = ai ; (4.9) pi because the production constraint tells us that F (x) = q. Plugging this into the production constraint give us q p1 a1 a1 p1 a1 an pn a1 an q pn an = q an a1 +:::+an a1 +:::+an q = q: But a1 + ::: + an = 1, so the above expression reduces further to a1 p1 a1 p1 a1 a1 an pn an pn an q = q an = 1 47 CHAPTER 4. CONSTRAINED OPTIMIZATION = p1 a1 a1 pn an an : We can substitute this back into (4.9) to get q x i = ai pi a pn ai p 1 1 = p i a1 an (4.10) an q: This is the input demand function, and it depends on the amount of output being produced (q), the input prices (p1 ; :::; pn ), and the exponents of the Cobb-Douglas production function. This doesn’t look particularly useful or intuitive. It can be, though. Plug it back into the original objective function p1 x1 + ::: + pn xn to get the cost function C(q) = p1 x1 + ::: + pn xn a a a1 p 1 1 an pn n q + ::: + pn = p1 p 1 a1 an pn a1 an p1 pn p1 = a1 q + ::: + an a1 an a1 a1 an pn p1 q = (a1 + ::: + an ) a1 an a a p1 1 pn n = q; a1 an p1 a1 a1 a1 pn an pn an an q an q where the last equality holds because a1 + ::: + an = 1. This one is pretty easy to remember. And it has a cool comparative statics result: a a pn n p1 1 dC(q) : (4.11) = dq a1 an Why is this cool? There are three reasons. First, dC(q)=dq is marginal cost, and q appears nowhere on the right-hand side. This means that the Cobb-Douglas production function gives us constant marginal cost. Second, compare the marginal cost function to the original production function: F (x) = xa11 p1 M C(q) = a1 xann a1 pn an an : CHAPTER 4. CONSTRAINED OPTIMIZATION 48 You can get the marginal cost function by replacing the xi ’s in the production function with the corresponding pi =ai . And third, remember how, at the end of the the last section on interpreting the Lagrange multiplier, we said that in a cost-minimization problem the Lagrange multiplier is just marginal cost? Compare equations (4.10) and (4.11). They are the same. I told you so. 4.6 Problems 1. Use the Lagrange multiplier method to solve the following problem: max 12x2 y 4 x;y s.t. 2x + 4y = 120 [Hint: You should be able to check your answer against the general version of the problem in Section 4.5.] 2. Solve the following problem: maxa;b 3 ln a + 2 ln b s.t. 12a + 14b = 400 3. Solve the following problem: minx;y 16x + y x1=4 y 3=4 = 1 4. Solve the following problem: max 3xy + 4x x;y s.t. 4x + 12y = 80 5. Solve the following problem: min 5x + 2y x;y s.t. 3x + 2xy = 80 CHAPTER 4. CONSTRAINED OPTIMIZATION 49 6. This is a lame but instructive problem. A farmer has 10 acres of land and uses it to grow corn. Pro…t from growing an acre of corn is given by (x) = 400x + 2x2 , where x is the number of acres of corn planted. So, the farmer’s problem is maxx 400x + 2x2 s.t. x = 10 (a) Find the …rst derivative of the pro…t function. Does its sign make sense? (b) Find the second derivative of the pro…t function. make sense? Does its sign (c) Set up the Lagrangian and use it to …nd the optimal value of x. (Hint: It had better be 10.) (d) Interpret the Lagrange multiplier. (e) Find the marginal value of an acre of land without using the Lagrange multiplier. (f) The second derivative of the pro…t function is positive. Does that mean that pro…t is minimized when x = 10? 7. Another lame but instructive problem: A …rm has the capacity to use 4 workers at a time. Each worker can produce 4 units per hour, each worker must p be paid $10 per hour, and the …rm’s revenue function is R(L) = 30 L, where L is the number of workers employedp(fractional workers are okay). The …rm’s pro…t function is (L) = 30 4L 10L. It must solve the problem p max 30 4L 10L L s.t. L = 4 (a) Find the …rst derivative of the pro…t function. Does its sign make sense? (b) Find the second derivative of the pro…t function. make sense? Does its sign (c) Set up the Lagrangian and use it to …nd the optimal value of L. [Hint: It had better be 4.] CHAPTER 4. CONSTRAINED OPTIMIZATION 50 (d) Interpret the Lagrange multiplier. (e) Find the marginal pro…t from a worker without using the Lagrange multiplier. (f) The second derivative of the pro…t function is negative. Does that mean pro…t is maximized when L = 4? 8. Here is the obligatory comparative statics problem. chooses x and y to maxx;y x y 1 s.t. px x + py y = M A consumer where px > 0 is the price of good x, py > 0 is the price of good y, M > 0 is income, and 0 < < 1. (a) Show that x = M=px and y = (1 )M=py . (b) Find @x =@M and @y =@M . Can you sign them? (c) Find @x =@px and @y =@px . Can you sign them? 9. This is the same as problem 2.11 but done using Lagrange multipliers. A …rm (Bilco) can use its manufacturing facility to make either widgets or gookeys. Both require labor only. The production function for widgets is W = 20w1=2 ; where w denotes labor devoted to widget production, and the production function for gookeys is G = 30g, where g denotes labor devoted to gookey production. The wage rate is $11 per unit of time, and the prices of widgets and gokeys are $9 and $3 per unit, repsectively. The manufacturing facility can accomodate 60 workers and no more. (a) Use a Lagrangian to determine how much of each product Bilco should produce per unit of time. (b) Interpret the Lagrange multiplier. CHAPTER 4. CONSTRAINED OPTIMIZATION 51 10. A farmer has a …xed amount F of fencing material that she can use to enclose a property. She does not yet have the property, but it will be a rectangle with length L and width W . Furthermore, state law dictates that every property must have a side smaller than S in length, and in this case S < F=4. [This last condition makes the constraint binding, and other than that you need not worry about it.] By convention, W is always the short side, so the state law dictates that W S. The farmer wants to use the fencing to enclose the largest possible area, and she also wants to obey the law. (a) Write down the farmer’s constrained maximization problem. [Hint: There should be two constraints.] (b) Write down the Lagrangian with two mulitpliers, one for each constraint, and solve the farmer’s problem. [Hint: The solution will be a function of F and S.] Please use as the second multiplier. (c) Which has a greater impact on the area the farmer can enclose, a marginal increase in S or a marginal increase in F ? Justify your answer. CHAPTER 5 Inequality constraints The previous chapter treated all constraints as equality constraints. Sometimes this is the right thing to do. For example, the …rm’s cost-minimization problem is to …nd the least-cost combination of inputs to produce a …xed amount of output, q. The constraint, then, is that output must be q, or, letting F (x1 ; :::; xn ) be the production function when the inputs are x1 ; :::; xn , the constraint is F (x1 ; :::; xn ) = q: Other times equality constraints are not the right thing to do. The consumer choice problem, for example, has the consumer choosing a commodity bundle to maximize utility, subject to the constraint that she does not spend more than her income. If the prices are p1 ; :::; pn , the goods are x1 ; :::; xn , and income is M , then the budget constraint is p1 x1 + ::: + pn xn M. It may be the case that the consumer spends her entire income, in which case the constraint would hold with equality. If she gets utility from saving, 52 CHAPTER 5. INEQUALITY CONSTRAINTS 53 though, she may not want to spend her entire income, in which case the budget constraint would not hold with equality. Firms have capacity constraints. When they build manufacturing facilities, the size of the facility places a constraint on the maximum output the …rm can produce. A capacity-constrained …rm’s problem, then, is to maximize pro…t subject to the constraint that output not exceed capacity, or q q. In the real world …rms often have excess capacity, which means that the capacity constraint does not hold with equality. Finally, economics often has implicit nonnegativity constraints. Firms cannot produce negative amounts by transforming outputs back into inputs. After all, it is di¢cult to turn a cake back into ‡our, baking powder, butter, salt, sugar, and unbroken eggs. Often we want to assume that we cannot consume negative amounts. As economists we must deal with these nonnegativity constraints. The goal for this chapter is to …gure out how to deal with inequality constraints. The best way to do this is through a series of exceptionally lame examples. What makes the examples lame is that the solutions are so transparent that it is hardly worth going through the math. The beauty of lame examples, though, is that this transparency allows you to see exactly what is going on. 5.1 Lame example - capacity constraints Let’s begin with a simple unconstrained pro…t maximization problem. The …rm chooses an amount to produce x, the market price is …xed at 80, and the cost function is 4x2 . The problem is max 80x x 4x2 . The …rst-order condition is 80 8x = 0, so the optimum is x = 10. The second-order condition is 8<0 which obviously holds, so the optimum is actually a maximum. The problem is illustrated in Figure 5.1. 54 CHAPTER 5. INEQUALITY CONSTRAINTS $ π(x) x 8 10 12 Figure 5.1: A lame example using capacity constraints 5.1.1 A binding constraint Now let’s put in a capacity constraint: x 8. This will obviously restrict the …rm’s output because it would like to produce 10 units but can only produce 8. (See why it’s a lame example?) The constraint will hold with equality, in which case we say that the constraint is binding. Let’s look at the math. max 80x 4x2 x s.t. x 8 Form the Lagrangian L(x; ) = 80x 4x2 + (8 x): The …rst-order conditions are @L = 80 8x =0 @x @L = 8 x = 0: @ The second equation tells us that x = 8, and the …rst equation tells us that = 80 64 = 16: 55 CHAPTER 5. INEQUALITY CONSTRAINTS So far we’ve done nothing new. The important step here is to think about . Remember that the interpretation of the Lagrange multiplier is that it is the marginal value of relaxing the constraint. In this case value is pro…t, so it is the marginal pro…t from relaxing the constraint. We can compute this directly: (x) = 80x 4x2 0 (x) = 80 8x 0 (8) = 80 64 = 16 Plugging the constrained value (x = 8) into the marginal pro…t function 0 (x) tells us that when output is 8, an increase in output leads to an increase in pro…t by 16. And this is exactly the Lagrange multiplier. 5.1.2 A nonbinding constraint Now let’s change the capacity constraint to x 12 and solve the problem intuitively. First, we know that pro…t reaches its unconstrained maximum when x = 10. The constraint does not rule this level of production out, so the constrained optimum is also x = 10. Because of this the capacity constraint is nonbinding, that is, it does not hold with equality. Nonbinding constraints are sometimes called slack. Let’s think about the Lagrange multiplier. We know that it is the marginal value of relaxing the constraint. How would pro…t change if we relaxed the constraint from x 12 to, say, x 13? The unconstrained maximum is still feasible, so the …rm would still produce 10 units and still generate exactly the same amount of pro…t. So, the marginal value of relaxing the constraint must be zero, and we have = 0. Now that we know the answers, let’s go back and look at the problem. 4x2 max 80x x s.t. x 12 This problem generates the Lagrangian L(x; ) = 80x 4x2 + (12 x): Since we already know the answers, let’s plug them in. know that = 0, so the Lagrangian becomes L(x; ) = 80x 4x2 In particular, we 56 CHAPTER 5. INEQUALITY CONSTRAINTS which is just the original unconstrained pro…t function. We arrived at our answers (x = 10, = 0) intuitively. How can we get them mechanically? After all, the purpose of the math is to make sure we get the answers right without relying solely on our intuition. One thing for sure is that we will need a new approach. To see why, suppose we analyze our 12-unit constraint problem in the usual way. Di¤erentiating the Lagrangian yields @L = 80 8x =0 @x @L = 12 x = 0 @ The second equation obviously implies that x = 12, in which case the …rst equation tells us that = 80 8x = 80 96 = 16. If we solve the problem using our old approach we …nd that (1) the constraint is binding, which is wrong, and (2) the Lagrange multiplier is negative, which means that relaxing the constraint makes pro…t even lower. You can see this in Figure 5.1. When output is 12 the pro…t function is downward-sloping. Since the Lagrange multiplier is marginal pro…t, we get a negative Lagrange multiplier when we are past the pro…t-maximizing level of output. 5.2 A new approach The key to the new approach is thinking about how the Lagrangian works. Suppose that the problem is max f (x1 ; :::; xn ) x1 ;:::;xn s.t. g(x1 ; :::; xn ) M The Lagrangian is L(x1 ; :::; xn ; ) = f (x1 ; :::; xn ) + [M g(x1 ; :::; xn )]: (5.1) When the constraint is binding, the term M g(x1 ; :::; xn ) is zero, in which case L(x1 ; :::; xn ; ) = f (x1 ; :::; xn ). When the constraint is nonbinding the Lagrange multiplier is zero, in which case L(x1 ; :::; xn ; ) = f (x1 ; :::; xn ) once again. So we need a condition that says [M g(x1 ; :::; xn )] = 0: 57 CHAPTER 5. INEQUALITY CONSTRAINTS Note that @[ [M g(x1 ; :::; xn )]] = M g(x1 ; :::; xn ) @ and in the old approach we set this equal to zero. We can no longer do this when the constraint is nonbinding, but notice that @[ [M g(x1 ; :::; xn )]] = (M @ g(x1 ; :::; xn )): This is exactly what we need to be equal to zero. We also need to restrict the Lagrange multiplier to be nonnegative. Remember from the lame example when the capacity constraint was binding at x = 12 we got a negative Lagrange multiplier, and that was the wrong answer. In fact, looking at expression (5.1) we can make L really large by making both and (M g(x1 ; :::; xn )) really negative. But when (M g(x1 ; :::; xn )) < 0 we have violated the constraint, so that is not allowed. The condition that @L = [(M @ g(x1 ; :::; xn )] = 0 is known as a complementary slackness condition. It says that one of two constraints must bind. One constraint is 0, and it binds if = 0, in which case the complementary slackness condition holds. The other constraint is g(x1 ; :::; xn ) M , and it binds if g(x1 ; :::; xn ) = M , in which case the complementary slackness condition also holds. If one of the constraints is slack, the other one has to bind. The beauty of the complementary slackness condition is that it forces one of two constraints to bind using a single equation. The …rst-order conditions for the inequality-constrained problem are @L = 0 @x1 .. . @L = 0 @xn @L = 0 @ 0 58 CHAPTER 5. INEQUALITY CONSTRAINTS The …rst set of conditions (@L=@xi = 0) are the same as in the standard case. The last two conditions are the ones that are di¤erent. The secondlast condition ( @L=@ = 0) guarantees that either the constraint binds, in which case @L=@ = 0, or the constraint does not bind, in which case = 0. The last condition says that the Lagrange multiplier cannot be negative, which means that relaxing the constraint cannot reduce the value of the objective function. We have a set of …rst-order conditions, but this does not tell us how to solve them. To do this, let’s go back to our lamest example: max 80x x s.t. x 4x2 12 which generates the Lagrangian L(x; ) = 80x 4x2 + (12 x): The …rst-order conditions are @L = 80 8x =0 @x @L = (12 x) = 0 @ 0 Now what? The methodology for solving this system is tedious, but it works. The second equation ( (12 x) = 0) is true if either (1) = 0, (2) 12 x = 0, or (3) both. So what we have to do is …nd the solution when = 0 and …nd the solution when 12 x = 0. Let’s see what happens. Case 1: = 0. If = 0 then the second and third conditions obviously hold. Plugging = 0 into the …rst equation yields 80 8x 0 = 0, or x = 10. Case 2: 12 x = 0. Then x = 12, and plugging this into the …rst equation yields = 80 96 = 16, which violates the last condition. So case 2 cannot be the answer. We are left with only one solution, and it is the correct one: x = 10, = 0. The general methodology for multiple constraints is as follows: Construct a Lagrange multiplier for each constraint. Each Lagrange multiplier can be 59 CHAPTER 5. INEQUALITY CONSTRAINTS either zero, in which case that constraint is nonbinding, or it can be positive, in which case its constraint is binding. Then try all possible combinations of zero/positive multipliers. Most of them will lead to violations. If only one does not lead to a violation, that is the answer. If several combinations do not lead to violations, then you must choose which one is the best. You can do this by plugging the values you …nd into the objective function. If you want to maximize the objective function, you choose the case that generates the highest value of the objective function. 5.3 Multiple inequality constraints Let’s look at another lame example: 1 2 max x 3 y 3 x;y s.t. x + y x+y 60 120 Why is this a lame example? Because we know that the second constraint must be nonbinding. After all if a number is smaller than 60, it must also be strictly smaller than 120. The solution to this problem will be the same as the solution to the problem 1 2 max x 3 y 3 x;y s.t. x + y 60 This looks like a utility maximization problem, as can be seen in Figure 5.2. The consumer’s utility function is u(x; y) = x1=3 y 2=3 , the prices of the two goods are px = py = 1, and the consumer has 60 to spend. The utility function is Cobb-Douglas, and from what we learned in Section 4.5 we know that x = 20 and y = 40. We want to solve it mechanically, though, to learn the steps. Assign a separate Lagrange multiplier to each constraint to get the Lagrangian L(x; y; 1; 2) 1 2 = x3 y 3 + 1 (60 x y) + 2 (120 x y): 60 CHAPTER 5. INEQUALITY CONSTRAINTS y 120 60 E 40 20 60 x 120 Figure 5.2: A lame example with two budget constraints The …rst-order conditions are @L @x @L @y @L 1 @ 1 @L 2 @ 2 1 2 1 y 2=3 3 x2=3 2 x1=3 = 3 y 1=3 = = 1 (60 = 2 (120 1 2 =0 1 2 =0 x x y) = 0 y) = 0 0 0 This time we have four possible cases: (1) 1 = 2 = 0, so that neither constraint binds. (2) 1 > 0, 2 = 0, so that only the …rst constraint binds. (3) 1 = 0, 2 > 0, so that only the second constraint binds. (4) 1 > 0, 2 > 0, so that both constraints bind. As before, we need to look at all four cases. Case 1: 1 = 2 = 0. In this case the …rst equation in the …rst-order conditions reduces to 31 (y=x)2=3 = 0, which implies that y = 0. The second equation reduces to 23 (x=y)1=3 = 0, but this cannot be true if y = 0 because we are not allowed to divide by zero. So Case 1 cannot be the answer. There CHAPTER 5. INEQUALITY CONSTRAINTS 61 is an easier way to see this, though. If neither constraint binds, the problem becomes 1 2 max x 3 y 3 x;y The objective function is increasing in both arguments, and since there are no constraints we want both x and y to be as large as possible. So x ! 1 and y ! 1. But this obviously violates the constraints. Case 2: 1 > 0, 2 = 0. The …rst-order conditions reduce to 1 y 2=3 3 x2=3 2 x1=3 3 y 1=3 60 x 1 = 0 1 = 0 y = 0 and the solution to this system is x = 20, y = 40, and 1 = 13 22=3 > 0. The remaining constraint, x + y 120, is satis…ed because x + y = 60 < 120. Case 2 works, and it corresponds to the case shown in Figure 5.2. Case 3: 1 = 0, 2 > 0. Now the …rst-order conditions reduce to 1 y 2=3 3 x2=3 2 x1=3 3 y 1=3 120 x 2 = 0 2 = 0 y = 0 The solution to this system is x = 40, y = 80, and 2 = 31 22=3 > 0. The remaining constraint, x + y 60, is violated because x + y = 120, so Case 3 does not work. There is an easier way to see this, though, just by looking at the constraints and exploiting the lameness of the example. If the second constraint binds, x + y = 120. But then the …rst constraint, x + y 60 cannot possibly hold. Case 4: 1 > 0, 2 > 0. The …rst of these conditions implies that the …rst constraint binds, so that x + y = 60. The second condition implies that x + y = 120, so that we are on the outer budget constraint in Figure 5.2. But then the inner budget constraint is violated, so Case 4 does not work. Only Case 2 works, so we know the solution: x = 20, y = 40, 1 = 1 2=3 2 , and 2 = 0. 3 62 CHAPTER 5. INEQUALITY CONSTRAINTS 5.4 A linear programming example A linear programming problem is one in which the objective function and all of the constraints are linear, such as in the following example: max 2x + y x;y s.t. 3x + 4y x y 60 0 0 This problem has three constraints, so we must use the multiple constraint methodology from the preceding section. It is useful in what follows to refer to the …rst constraint, 3x + 4y 60, as a budget constraint. The other two constraints are nonnegativity constraints. The Lagrangian is L(x; y; 1; 2; 3) = 2x + y + 1 (60 3x 4y) + 2 (x 0) + 3 (y 0): Notice that we wrote all of the constraint terms, that is (60 3x 4y) and (x 0) and (y 0) so that they are nonnegative. We have been doing this throughout this book. The …rst-order conditions are @L @x @L @y @L 1 @ 1 @L 2 @ 2 @L 3 @ 3 1 = 2 3 1 + 2 =0 = 1 4 1 + 3 =0 = 1 (60 = 2x =0 = 3y =0 0, 2 3x 0, 4y) = 0 3 0 Since there are three constraints, there are 23 = 8 possible cases. We are going to narrow some of them down intuitively before going on. The …rst CHAPTER 5. INEQUALITY CONSTRAINTS 63 constraint is like a budget constraint, and the objective function is increasing in both of its arguments. The other two constraints are nonnegativity constraints, saying that the consumer cannot consume negative amounts of the goods. Since there is only one budget-type constraint, it has to bind, which means that 1 > 0. The only question is whether one of the other two constraints binds. A binding budget constraint means that we cannot have both x = 0 and y = 0, because if we did then we would have 3x + 4y = 0 < 60, and the budget constraint would not bind. We are now left with three possibilities: (1) 1 > 0 so the budget constraint binds, 2 > 0, and 3 = 0, so that x = 0 but y > 0. (2) 1 > 0, 2 = 0, and 3 > 0, so that x > 0 and y = 0. (3) We will 1 > 0, 2 = 0, and 3 = 0, so that both x and y are positive. consider these one at a time. Case 1: 1 > 0, 2 > 0, 3 = 0. Since 2 > 0 the constraint x 0 must bind, so x = 0. For the budget constraint to hold we must have y = 15. This yields a value for the objective function of 2x + y = 2 0 + 15 = 15: Case 2: 1 > 0, 2 = 0, 3 > 0. This time we have y = 0, and the budget constraint implies that x = 20. The objective function then takes the value 2x + y = 2 20 + 0 = 40: Case 3: 1 > 0, 2 = 0, 3 = 0. In this case both x and y are positive. The …rst two equations in the …rst-order conditions become 2 1 3 4 1 1 = 0 = 0 The …rst of these reduces to 1 = 2=3, and the second reduces to 1 = 1=4. These cannot both hold, so Case 3 does not work. We got solutions in Case 1 and Case 2, but not in Case 3. So which is the answer? The one that generates a larger value for the objective function. In Case 1 the maximized objective function took a value of 15 and in Case 2 it took a value of 40, which is obviously higher. So Case 2 is the solution. To see what we just did, look at Figure 5.3. The three constraints identify a triangular feasible set. Case 1 is the corner solution where the budget line meets the y-axis, and Case 2 is the corner solution where the budget line 64 CHAPTER 5. INEQUALITY CONSTRAINTS y 15 indifference curve x 20 Figure 5.3: Linear programming problem meets the x-axis. Case 3 is an "interior" solution that is on the budget line but not on either axis. The objective was to …nd the point in the feasible set that maximized the function f (x; y) = 2x + y. We did this by comparing the values we got at the two corner solutions, and we chose the corner solution that gave us the larger value. The methodology for solving linear programming problems involves …nding all of the corners and choosing the corner that yields the largest value of the objective function. A typical problem has more than two dimensions, so it involves …nding x1 ; :::; xn , and it has more than one budget constraint. This generates lots and lots of corners, and the real issue in the study of linear programming is …nding an algorithm that e¢ciently checks the corners. 5.5 Kuhn-Tucker conditions Economists sometimes structure the …rst-order conditions for inequalityconstrained optimization problems di¤erently than the way we have done it so far. The alternative formulation was developed by Harold Kuhn and A.W. Tucker, and it is known as the Kuhn-Tucker formulation. The …rstorder conditions we will derive are known as the Kuhn-Tucker conditions. Begin with the very general maximization problem, letting x be the vector 65 CHAPTER 5. INEQUALITY CONSTRAINTS x = (x1 ; :::; xn ): max f (x) x s.t. g 1 (x) x1 b1 ; :::; g k (x) bk 0; :::; xn 0: There are k "budget-type" constraints and n non-negativity constraints. To solve this, form the Lagrangian L(x; 1 ; :::; k ; v1 ; :::; vk ) = f (x) + k X i [bi i=1 g i (x)] + n X vj xj : j=1 We get the following …rst-order conditions: @L @f @g 1 = ::: 1 @x1 @x1 @x1 .. . @f @g 1 @L = ::: 1 @xn @xn @xn @L = 1 [b1 g 1 (x)] = 0 1 @ 1 .. . @L = k [bk g k (x)] = 0 k @ k @L v1 = v1 x1 = 0 @v1 .. . @L = vn xn = 0 vn @vn 1 ; :::; k ; v1 ; :::; vn @g k + v1 = 0 k @x1 k @g k + vn = 0 @xn 0 There are 2n + k conditions plus the n + k nonnegativity constraints for the 66 CHAPTER 5. INEQUALITY CONSTRAINTS multipliers. It is useful to have some shorthand to shrink this system down: @L = 0 for @xi @L = 0 for j @ j vi xi = 0 for 0 for j vi 0 for i = 1; :::; n (5.2) j = 1; :::; k i = 1; :::; n j = 1; :::; k i = 1; :::; n Suppose instead we had constructed a di¤erent Lagrangian: K(x; 1 ; :::; k) = f (x) + k X i [bi g i (x)]: i=1 This Lagrangian, known as the Kuhn-Tucker Lagrangian, only has k multipliers for the k budget-type constraints, and no multipliers for the nonnegativity constraints. The two Lagrangians are related, with L(x; 1 ; :::; k ; v1 ; :::; vk ) = K(x; 1 ; :::; k) + n X vj xj : j=1 We can rewrite the system of …rst-order conditions (5.2) as @K @L = + vi = 0 for i = 1; :::; n @xi @xi @L @K = j = 0 for j = 1; :::; k j @ j @ j vi xi = 0 for i = 1; :::; n 0 for j = 1; :::; k j vi 0 for i = 1; :::; n Pay close attention to the …rst and third equations. If vi = 0 then the …rst equation yields @K @K = 0 =) xi = 0: vi = 0 =) @xi @xi On the other hand, if vi > 0 then the i-th inequality constraint, xi 0, is binding which means that vi > 0 =) xi = 0 =) xi @K = 0: @xi 67 CHAPTER 5. INEQUALITY CONSTRAINTS Either way we have @K = 0 for i = 1; :::; n. @xi The Kuhn-Tucker conditions use this information. …rst-order conditions is xi @K @xi @K j @ j xi The new set of = 0 for i = 1; :::; n (5.3) = 0 for j = 1; :::; k 0 for j = 1; :::; k 0 for i = 1; :::; n j xi This is a system of n + k equations in n + k unknowns plus n + k nonnegativity constraints. Thus, it simpli…es the original set of conditions by removing n equations and n unknowns. It is also a very symmetriclooking set of conditions. Remember that the Kuhn-Tucker Lagrangian is K(x1 ; :::; xn ; 1 ; :::; k ). Instead of distinguishing between x’s and ’s, let them all be z’s, in which case the Kuhn-Tucker Lagrangian is K(z1 ; :::; zn+k ). Then the Kuhn-Tucker conditions reduce to zj @K=@zj = 0 and zj 0 for j = 1; :::; n+k. This is fairly easy to remember, which is an advantage. The key to Kuhn-Tucker conditions, though, is remembering that they are just a particular reformulation of the standard inequality-constrained optimization problem with multiple constraints. 5.6 Problems 1. Consider the following problem: maxx;y x2 y s.t. 2x + 3y 24 4x + y 20 The Lagrangian can be written L(x; y; 1; 2) = x2 y + 1 (24 2x 3y) + 2 (20 4x y) 68 CHAPTER 5. INEQUALITY CONSTRAINTS (a) Solve the alternative problem maxx;y x2 y s.t. 2x + 3y = 24 Do the resulting values of x and y satisfy 4x + y 20? (b) Solve the alternative problem maxx;y x2 y s.t. 4x + y = 20 Do the resulting values of x and y satisfy 2x + 3y 24? (c) Based on your answers to (a) and (b), which of the two constraints bind? What do these imply about the values of 1 and 2 ? (d) Solve the original problem. (e) Draw a graph showing what is going on in this problem. 2. Consider the following problem: maxx;y x2 y s.t. 2x + 3y 24 4x + y 36 The Lagrangian can be written L(x; y; 1; 2) = x2 y + 1 (24 2x 3y) + 2 (36 4x (a) Solve the alternative problem maxx;y x2 y s.t. 2x + 3y = 24 Do the resulting values of x and y satisfy 4x + y 36? (b) Solve the alternative problem maxx;y x2 y s.t. 4x + y = 36 Do the resulting values of x and y satisfy 2x + 3y 24? y) 69 CHAPTER 5. INEQUALITY CONSTRAINTS (c) Based on your answers to (a) and (b), which of the two constraints bind? What do these imply about the values of 1 and 2 ? (d) Solve the original problem. (e) Draw a graph showing what is going on in this problem. 3. Consider the following problem: max 4xy x;y 3x2 s.t. x + 4y 5x + 2y 36 45 The Lagrangian can be written L(x; y; 1; 2) = 4xy 3x2 + 1 (36 x 4y) + 2 (45 5x 2y) (a) Solve the alternative problem 3x2 max 4xy x;y s.t. x + 4y = 36 Do the resulting values of x and y satisfy 5x + 2y (b) Solve the alternative problem 45? 3x2 max 4xy x;y s.t. 5x + 2y = 45 Do the resulting values of x and y satisfy x + 4y 36? (c) Find the solution to the original problem, including the values of 1 and 2 . 4. Consider the following problem: max 3xy x;y 8x s.t. x + 4y 5x + 2y 24 30 The Lagrangian can be written L(x; y; 1; 2) = 3xy 8x + 1 (24 x 4y) + 2 (30 5x 2y) 70 CHAPTER 5. INEQUALITY CONSTRAINTS (a) Solve the alternative problem max 3xy x;y 8x s.t. x + 4y = 24 Do the resulting values of x and y satisfy 5x + 2y 30? (b) Solve the alternative problem max 3xy x;y 8x s.t. 5x + 2y = 30 Do the resulting values of x and y satisfy x + 4y 24? (c) Find the solution to the original problem, including the values of 1 and 2 . 5. Consider the following problem: maxx;y x2 y s.t. 4x + 2y 42 x 0 y 0 (a) Write down the Kuhn-Tucker Lagrangian for this problem. (b) Write down the Kuhn-Tucker conditions. (c) Solve the problem. 6. Consider the following problem: max xy + 40x + 60y x;y s.t. x + y x; y 12 0 (a) Write down the Kuhn-Tucker Lagrangian for this problem. (b) Write down the Kuhn-Tucker conditions. (c) Solve the problem. PART II SOLVING SYSTEMS OF EQUATIONS (linear algebra) CHAPTER 6 Matrices Matrices are 2-dimensional arrays of numbers, and they are useful for many things. They also behave di¤erently that ordinary real numbers. This chapter tells how to work with matrices and what they are for. 6.1 Matrix algebra A matrix is a rectangular array of numbers, such as the one below: 1 0 6 5 3 A: A=@ 2 1 4 Matrices are typically denoted by capital letters. They have dimensions corresponding to the number of rows and number of columns. The matrix A above has 3 rows and 2 columns, so it is a 3 2 matrix. Matrix dimensions are always written as (# rows) (# columns). An element of a matrix is one of the entries. The element in row i and 72 73 CHAPTER 6. MATRICES column j is denoted aij , and so in 0 a11 B a21 B A = B .. @ . an1 general a matrix looks like 1 a12 a1k a22 a2k C C .. .. C : .. . . . A an2 ank The matrix A above is an n k matrix. A matrix in which the number of rows equals the number of columns is called a square matrix. In such a matrix, elements of the form aii are called diagonal elements because they land on the diagonal of the square matrix. An n-dimensional vector can be thought of as an n 1 matrix. Therefore, in matrix notation vectors are written vertically: 1 0 x1 C B x = @ ... A : xn When we write a vector as a column matrix we typically leave o¤ the accent and write it simply as x. Matrix addition is done element by element: 1 1 0 1 0 0 a11 + b11 a1k + b1k b11 b1k a11 a1k C B .. .. .. . C B .. C + B .. . . ... ... A: @ . . .. A = @ . . . A @ . an1 + bn1 ank + bnk bn1 bnk an1 ank Before one can add matrices, though, it is important to make sure that the dimensions of the two matrices are identical. In the above example, both matrices are n k. Just like with vectors, it is possible to multiply a matrix by a scalar. This is done element by element: 0 1 0 1 a11 a1k ta11 ta1k B .. C = B .. .. C : ... .. tA = t @ ... . . A @ . . A an1 ank tan1 tank The big deal in matrix algebra is matrix multiplication. To multiply matrices A and B, several things are important. First, the order matters, as 74 CHAPTER 6. MATRICES you will see. Second, the number of columns in the …rst matrix must equal the number of rows in the second. So, one must multiply an n k matrix on the left by a k m matrix on the right. The result will be an n m matrix, with the k’s canceling out. The formula for multiplying matrices is as follows. Let C = AB, with A an n k matrix and B a k m matrix. Then k X cij = ais bsj : s=1 This is easier to see when 0 a11 B a21 B C = AB = B .. @ . an1 Element c11 is we write the matrices A 10 b11 a12 a1k B C a22 a2k C B b21 .. C B .. .. .. . . A@ . . bk1 an2 ank and B side-by-side: 1 b12 b1m b22 b2m C C .. C : .. . . . . A . bk2 bkm c11 = a11 b11 + a12 b21 + ::: + a1s bs1 + ::: + a1k bk1 : So, element c11 is found by multiplying each member of row 1 in matrix A by the corresponding member of column 1 in matrix B and then summing. Element cij is found by multiplying each member of row i in matrix A by the corresponding member of column j in matrix B and then summing. For there to be the right number of elements for this to work, the number of columns in A must equal the number of rows in B. As an example, multiply the two matrices below: A= 6 4 1 3 ,B= 3 4 2 4 : Then AB = 6 3 + ( 1)( 2) 6 4 + ( 1)4 4 3 + 3( 2) 4 4+3 4 = 20 20 6 28 : 3 6+4 4 3( 1) + 4 3 ( 2)6 + 4 4 ( 2)( 1) + 4 3 = 34 9 4 14 : However, BA = 75 CHAPTER 6. MATRICES Obviously, AB 6= BA, and matrix multiplication is not commutative. Because of this, we use the terminology that we left-multiply by B when we want BA and right-multiply by B when we want AB. The square matrix 0 1 1 0 0 B 0 1 0 C B C I = B .. .. . . .. C @ . . . . A 0 0 1 is special, and is called the identity matrix. To see why it is special, consider any n k matrix A, and let I be the n-dimensional identity matrix. Letting B = IA, we get bij = 0 a1j + 0 a2j + ::: + 1 aij + ::: + 0 anj = aij . So, IA = A. The same thing happens when we right-multiply A by a kdimensional identity matrix. Then AI = A. So, multiplying a matrix by the identity matrix is the same as multiplying an ordinary number by 1. The transpose of the matrix 1 0 a11 a12 a1k B a21 a22 a2k C C B A = B .. .. C : .. . . @ . . . A . an1 an2 ank is the matrix AT given by 0 B B AT = B @ a11 a21 a12 a22 .. .. . . a1k a2k an1 an2 .. .. . . ank 1 C C C: A The transpose is generated by switching the rows and columns of the original matrix. Because of this, the transpose of an n k matrix is a k n matrix. Note that (AB)T = B T AT so that the transpose of the product of two matrices is the product of the transposes of the two matrices, but you have to switch the order of the matrices. To check this, consider the following example employing the same two matrices we used above. A= 6 4 1 3 ,B= 3 4 2 4 : 76 CHAPTER 6. MATRICES AB = 20 20 6 28 6 4 1 3 AT = B T AT = 3 4 2 4 20 6 20 28 , (AB)T = , BT = 6 4 1 3 3 4 2 4 : : = 20 6 20 28 = (AB)T ; = 34 4 9 14 = (BA)T : as desired, but AT B T = 6.2 6 4 1 3 3 4 2 4 Uses of matrices Suppose that you have a system of n equations in n unknowns, such as this one: a11 x1 + ::: + a1n xn = b1 a21 x1 + ::: + a2n xn = b2 (6.1) .. . an1 x1 + ::: + ann xn = bn We can write this system easily using matrix notation. Letting 1 1 0 1 0 0 b1 x1 a11 a1n B .. C , x = B .. C , and b = B .. C , .. A = @ ... @ . A @ . A . . A bn xn an1 ann we can rewrite the system of equations (6.1) as Ax = b: (6.2) The primary use of matrices is to solve systems of equations. As you have seen in the optimization examples, systems of equations regularly arise in economics. Equation (6.2) raises two questions. First, when does a solution exist, that is, when can we …nd a vector x such that Ax = b? Second, how do we …nd the solution when it exists? The answers to both questions depend on the inverse matrix A 1 , which is a matrix having the property that A 1 A = AA 1 = I, 77 CHAPTER 6. MATRICES that is, the matrix that you multiply A by to get the identity matrix. Remembering that the identity matrix plays the role of the number 1 in matrix multiplication, and that for ordinary numbers the inverse of y is the number y 1 such that y 1 y = 1, this formula is exactly what we are looking for. If A has an inverse (and that is a big if), then (6.2) can be solved by left-multiplying both sides of the equation by the inverse matrix A 1 : A 1 Ax = A 1 b x = A 1b For ordinary real numbers, every number except 0 has a multiplicative inverse. Many more matrices than just one fail to have an inverse, though, so we must devote considerable attention to whether or not a matrix has an inverse. A second use of matrices is for deriving second-order conditions for multivariable optimization problems. Recall that in a single-variable optimization problem with objective function f , the second-order condition is determined by the sign of the second derivative f 00 . When f is a function of several variables, however, we can write the vector of …rst partial derivatives 0 @f 1 @x1 B .. C @ . A @f @xn and the matrix of second partials 0 @2f @2f B B B B @ @x21 @2f @x2 @x1 .. . @2f @xn @x1 @x1 @x2 @2f @x22 .. . @2f @xn @x2 .. @2f @x1 @xn @2f @x2 @xn . .. . @2f @x2n 1 C C C: C A The relevant second-order conditions will come from conditions on the matrix of second partials, and we will do this in Chapter 9. 6.3 Determinants Determinants of matrices are useful for determining (hence the name) whether a matrix has an inverse and also for solving equations such as (6.2). The de- 78 CHAPTER 6. MATRICES terminant of a square matrix A (and the matrix must be square) is denoted jAj. De…ning it depends on the size of the matrix. Start with a 1 1 matrix A = (a11 ). The determinant jAj is simply a11 . It can be either positive or negative, so don’t confuse the determinant with the absolute value, even though they both use the same symbol. Now look at a 2 2 matrix A= a11 a12 a21 a22 : Here the determinant is de…ned as jAj = For a 3 matrix a11 a12 a21 a22 = a11 a22 a21 a12 : 3 matrix we go through some more steps. Begin with the 3 1 0 a11 a12 a13 A = @ a21 a22 a23 A : a31 a32 a33 3 We can get a submatrix of A by deleting a row and column. For example, if we delete the second row and the …rst column we are left with the submatrix A21 = a12 a13 a32 a33 : In general, submatrix Aij is obtained from A by deleting row i and column j. Note that there is one submatrix for each element, and you can get that submatrix by eliminating the element’s row and column from the original matrix. Every element also has something called a cofactor which is based on the element’s submatrix. Speci…cally, the cofactor of aij is the number cij given by cij = ( 1)i+j jAij j, that is, it is the determinant of the submatrix Aij multiplied by 1 if i + j is odd and multiplied by 1 if i + j is even. Using these de…nitions we can …nally get the determinant of a 3 3 matrix, or any other square matrix for that matter. There are two ways to do it. The most common is to choose a column j. Then jAj = a1j c1j + a2j c2j + ::: + anj cnj . 79 CHAPTER 6. MATRICES Before we see what this means for a 3 3 matrix let’s check that it works for a 2 2 matrix. Choosing column j = 1 gives us a11 a12 a21 a22 = a11 c11 + a12 c12 = a11 a22 + a12 ( a21 ); where c11 = a22 because 1 + 1 is even and c21 = a12 because 1 + 2 is odd. We get exactly the same thing if we choose the second column, j = 2: a11 a12 a21 a22 = a12 c12 + a22 c22 = a12 ( a21 ) + a22 a11 : Finally, let’s look at a 3 3 matrix, choosing j = 1. a11 a12 a13 a21 a22 a23 a31 a32 a33 = a11 c11 + a21 c21 + a31 c31 = a11 (a22 a33 a32 a23 ) a21 (a12 a33 a32 a13 ) +a31 (a12 a23 a22 a13 ): We can also …nd determinants by choosing a row. then the determinant is given by If we choose row i, jAj = ai1 ci1 + ai2 ci2 + ::: + ain cin . The freedom to choose any row or column allows one to use zeros strategically. For example, when evaluating the determinant 6 8 2 0 9 4 1 0 7 it would be best to choose the second row because it has two zeros, and the determinant is simply a21 c21 = 2( 60) = 120: 6.4 Cramer’s rule The process of using determinants to solve the system of equations given by Ax = b is known as Cramer’s rule. Begin with the matrix 1 0 a11 a1n B .. C : ... A = @ ... . A an1 ann 80 CHAPTER 6. MATRICES Construct the matrix Bi from A by replacing column vector b, so that 0 b1 a12 B b2 a22 B B1 = B .. .. ... @ . . bn an2 and 0 B B Bi = B @ a11 a21 .. . a1(i a2(i .. . ... an1 an(i 1) 1) 1) the i-th column of A with the a1n a2n .. . ann 1 C C C A b1 a1(i+1) b2 a2(i+1) .. .. . . bn an(i+1) .. a1n a2n .. . . ann 1 C C C: A According to Cramer’s rule, the solution to Ax = b is the column vector x where jBi j xi = : jAj Let’s make sure this works using a simple example. equations is The system of 4x1 + 3x2 = 18 5x1 3x2 = 9 Adding the two equations together yields 9x1 = 27 x1 = 3 x2 = 2 Now let’s do it using Cramer’s rule. We have 4 5 A= 3 3 and b = 18 9 : Generate the matrices B1 = 18 9 3 3 and B2 = 4 18 5 9 : 81 CHAPTER 6. MATRICES Now compute determinants to get jAj = 27, jB1 j = 81, and jB2 j = 54: Applying Cramer’s rule we get x= 6.5 jB1 j jAj jB2 j jAj ! 81 27 54 27 = = 3 2 : Inverses of matrices One important implication of Cramer’s rule links the determinant of A to the existence of an inverse. To see why, recall that the solution, if it exists, to the system Ax = b is x = A 1 b. Also, we know from Cramer’s rule that xi = jBi j = jAj. For this number to exist, it must be the case that jAj = 6 0. This is su¢ciently important that it has a name: the matrix A is singular if jAj = 0 and it is nonsingular if jAj = 6 0. Singular matrices do not have inverses, but nonsingular matrices do. With some clever manipulation we can use Cramer’s rule to invert the matrix A. To see how, recall that the inverse is de…ned so that AA 1 = I. We want to …nd A 1 , and it helps to de…ne 1 0 x11 x1n B .. C ... A 1 = @ ... . A xn1 xnn and 0 1 x1i B C xi = @ ... A : xni Remember that the coordinate vector ei has zeros everywhere except for the i-th element, which is 1, and so 0 1 1 B 0 C B C e1 = B .. C @ . A 0 82 CHAPTER 6. MATRICES and ei is the column vector with eii = 1 and eij = 0 when j 6= i. Note that ei is the i-th column of the identity matrix. Then i-th column of the inverse matrix A 1 can be found by applying Cramer’s rule to the system Axi = ei . (6.3) Construct the matrix Bji from A by replacing the j-th column of A with the column vector ei . Then 0 1 a11 a1(j 1) 0 a1(j+1) a1n .. .. .. .. .. .. .. B C . . . . . . . B C B C a(i 1)(j 1) 0 a(i 1)(j+1) a(i 1)n C B a(i 1)1 B C ai(j 1) 1 ai(j+1) ain C Bji = B ai1 B C a(i+1)(j 1) 0 a(i+1)(j+1) a(i+1)n C B a(i+1)1 B C .. .. .. .. .. .. .. @ A . . . . . . . an1 an(j 1) 0 an(j+1) ann The solution to (6.3) is Bji : xji = jAj Once again we can only get an inverse if jAj = 6 0. We can simplify this further. Note that only one element of the j-th column of Bji is non-zero. So, we can take the determinant of Bji by using the j-th column and getting Bji = ( 1)i+j jAij j = cij , which is a cofactor of the matrix A. So, we get the formula for the inverse: xji = ( 1)i+j jAij j : jAj Note that the subscript on x is ji but the subscript on A is ij. Let’s check to see if this works for a 2 2 matrix. Use the matrix A= a b c d : We get B11 = 1 b 0 d , B12 = 0 b 1 d , B21 = a 1 c 0 , and B22 = a 0 c 1 : 83 CHAPTER 6. MATRICES The determinants are B11 = d, B12 = We also know that jAj = ad A 1 b, B21 = c, and B22 = a. bc. Thus, 1 = ad d c bc b a : We can check this easily: A 1A = = = 6.6 1 ad d c bc 1 ad bc 1 0 0 1 b a ad bc ca + ac : Problems 1. Perform the following computations: (a) 4 6 3 (b) 2 1 1 4 0 2 (c) @ 3 1 (d) 2 4 2 3 9 5 1 0 2 3 3 2 1 4 4 1 0 1 2 2 1 0 0 B 3 3 2 1 0 AB @ 7 2 1 0 7 4 5 10 0 3 1 1 1 1 @ 2 1 1 A@ 1 2 2 6 5 1 1 0 C C 1 A 0 1 5 1 A 1 2. Perform the following computations: 1 0 1 0 3 3 8 10 4 A (a) 6 @ 4 5 A 21 @ 6 1 2 6 2 a b c d ba + ba cb + ad 84 CHAPTER 6. MATRICES 10 0 3 A @ 1 6 2 1 0 5 B 6 1 3 0 B 0 1 5 2 1 @ 1 0 0 2 0 1 10 5 1 @ 1 3 1 5 0 2 0 5 1 @ 4 0 (b) 1 2 (c) (d) 1 4 5 A 5 1 1 4 2 4 C C 3 4 A 6 4 1 10 5 A@ 1 A 3 3. Find the determinants of the following matrices: (a) 0 2 1 4 5 1 0 2 A 6 3 1 (b) @ 2 7 2 0 4. Find the determinants of the following matrices: (a) 0 3 4 2 0 @ 1 3 (b) 0 6 6 1 1 1 0 A 1 5. Use Cramer’s rule to solve the following system of equations: 6x 2y 3z = 1 2x + 4y + z = 2 3x z = 8 6. Use Cramer’s rule to solve the following system of equations: 5x 2y + z = 9 3x y = 9 3y + 2z = 15 CHAPTER 6. MATRICES 7. Invert the following matrices: 2 3 2 1 (a) 1 1 1 2 (b) @ 0 1 1 A 1 1 0 0 8. Invert the following matrices: 4 2 (a) 0 5 1 @ 0 2 (b) 0 1 1 4 1 1 1 A 3 85 CHAPTER 7 Systems of equations Think about the general system of equations Ax = b where A is an n n matrix and x and b are n the system of equations (7.1) 1 vectors. This expands to a11 x1 + a12 x2 + ::: + a1n xn = b1 a21 x1 + a22 x2 + ::: + a2n xn = b2 .. . an1 x1 + an2 x2 + ::: + ann xn = bn The task for this chapter is to determine (1) whether the system has a solution (x1 ; :::; xn ), and (2) whether that solution is unique. We will use three examples to motivate our results. They use n = 2 to allow graphical analysis. 86 87 CHAPTER 7. SYSTEMS OF EQUATIONS Example 3 2x + y = 6 x y = 3 The solution to this one is (x; y) = (1; 4). Example 4 x 4y 2y = 1 2x = 2 This one has an in…nite number of solutions de…ned by (x; y) = (x; x 2 1 ). Example 5 x 2y y = 4 2x = 5 This one has no solution. 7.1 7.1.1 Identifying the number of solutions The inverse approach If A has an inverse A 1 , then left-multiplying both sides of (7.1) by A 1 yields x = A 1 b. So, what we really want to know is, when does an inverse exist? We already know that an inverse exists if the determinant is nonzero, that is, if jAj = 6 0. 7.1.2 Row-echelon decomposition Write the augmented matrix 0 a11 B .. B = (Ajb) = @ . an1 ... a1n .. . ann 1 b1 .. C . A bn which is n (n+1). The goal is to transform the matrix B through operations that consist of multiplying one row by a scalar and adding it to another row, and ending up with a matrix in which all the elements below the diagonal are 0. This is the row-echelon form of the matrix. 88 CHAPTER 7. SYSTEMS OF EQUATIONS Example 6 (3 continued) Form the augmented matrix B= Multiply the …rst row by R= 1 2 2 1 2 1 1 1 and add it to the second row to get 1 1 6 3 1 6 3 3 1 2 = 2 0 1 3 2 6 6 Example 7 (4 continued) Form the augmented matrix B= 1 2 2 4 1 2 Multiply the …rst row by 2 and add it to the second row to get R= 1 2 2+2 4 4 1 2+2 1 0 = 2 0 1 0 Example 8 (5 continued) Form the augmented matrix B= 1 2 1 2 4 5 Multiply the …rst row by 2 and add it to the second row to get R= 1 1 2+2 2 2 4 5+8 = 1 0 1 0 4 13 Example 3 has a unique solution, Example 4 has an in…nite number of them, and Example 5 has no solution. These results correspond to properties of the row-echelon matrix R. If the row-echelon form of the augmented matrix has only nonzero diagonal elements, there is a unique solution. If it has some rows that are zero, there are in…nitely many solutions. If there are rows with zeros everywhere except in the last column, there is no solution. De…nition 1 The rank of a matrix is the number of nonzero rows in its row-echelon form: Proposition 9 The n rank is n. n square matrix A has an inverse if and only if its 89 CHAPTER 7. SYSTEMS OF EQUATIONS y y y x – y = –3 x 2x + y = 6 2y – 2x = 5 x x – 2y = 1 4y – 2x = –2 x x–y=4 Figure 7.1: Graphing in (x; y) space: One solution when the lines intersect (left graph), in…nite number of solutions when the lines coincide (center graph), and no solutions when the lines are parallel (right graph) 7.1.3 Graphing in (x,y) space This is pretty simple and is shown in Figure 7.1. The equations are lines. In example 3 the equations constitute two di¤erent lines that cross at a single point. In example 4 the lines coincide. In example 5 the lines are parallel. We get a unique solution in example 3, an in…nite number of them in example 4, and no solution in example 5. What happens if we move to three dimensions? An equation reduces the dimension by one, so each equation identi…es a plane. Two planes intersect in a line. The intersection of a line and a plane is a point. So, we get a unique solution if the planes intersect in a single point, an in…nite number of solutions if the three planes intersect in a line or in a plane, and no solution if two or more of the planes are parallel. 7.1.4 Graphing in column space This approach is completely di¤erent but really useful. Each column of A is an n 1 vector. So is b. The question becomes, is b in the column space, that is, the space spanned by the columns of A? A linear combination of vectors a and b is given by xa + y b, where x and y are scalars. The span of the vectors x and y is the set of linear combinations of the 90 CHAPTER 7. SYSTEMS OF EQUATIONS y y y b = (4,5) (–2,4) (–1,2) (2,1) x (1,-1) b = (6,–3) x b = (1,–2) x (1,–2) Figure 7.2: Graphing in the column space: When the vector b is in the column space (left graph and center graph) a solution exists, but when the vector b is not in the column space there is no solution (right graph) two vectors. Example 9 (3 continued) The column vectors are (2; 1) and (1; 1). These span the entire plane. For any vector (z1 ; z2 ), it is possible to …nd numbers x and y that solve x(2; 1) + y(1; 1) = (z1 ; z2 ). Written in matrix notation this is z1 x 2 1 = : z2 y 1 1 We already know that the matrix has an inverse, so we can solve this. Look at left graph in Figure 7.2. The column space is spanned by the two vectors (2; 1) and (1; 1). The vector b = (6; 3) lies in the span of the column vectors. This leads to our rule for a solution: If b is in the span of the columns of A, there is a solution. Example 10 (4 continued) In the center graph in Figure 7.2, the column vectors are (1; 2) and ( 2; 4). They are on the same line, so they only span that line. In this case the vector b = (1; 2) is also on that line, so there is a solution. Example 11 (5 continued) In the right panel of Figure 7.2, the column vectors are (1; 2) and ( 1; 2), which span a single line. This time, though, the vector b = (4; 5) is not on that line, and there is no solution. 91 CHAPTER 7. SYSTEMS OF EQUATIONS Two vectors a and b are linearly dependent if there exists a scalar r 6= 0 such that a = rb. Two vectors a and b are linearly independent if there is no scalar r 6= 0 such that a = rb. Equivalently, a and b are linearly independent if there do not exist scalars r1 and r2 , not both equal to zero, such that r1 a + r2 b = 0: (7.2) One more way of writing this is that a and b are linearly independent if the only solution to the above equation has r1 = r2 = 0. This last one has some impact when we write it in matrix form. Suppose that the two vectors are 2-dimensional, and construct the matrix A by using the vectors a and b as its columns: a1 b 1 a2 b 2 A= : Now we can write (7.2) as A r1 r2 = 0 0 : If A has an inverse, there is a unique solution to this equation, given by r1 r2 =A 1 0 0 = 0 0 : So, invertibility of A and the linear independence of its columns are inextricably linked. Proposition 10 The square matrix A has an inverse if its columns are mutually linearly independent. 7.2 Summary of results The system of n linear equations in n unknowns given by Ax = b has a unique solution if: CHAPTER 7. SYSTEMS OF EQUATIONS 1. A 1 92 exists. 2. jAj = 6 0. 3. The row-echelon form of A has no rows with all zeros. 4. A has rank n. 5. The columns of A span n-dimensional space. 6. The columns of A are mutually linearly independent. The system of n linear equations in n unknowns given by Ax = b has an in…nite number of solutions if: 1. The row-echelon form of the augmented matrix (Ajb) has rows with all zeros. 2. The vector b is contained in the span of a subset of the columns of A. The system of n linear equations in n unknowns given by Ax = b has no solution if: 1. The row-echelon form of the augmented matrix (Ajb) has at least one row with all zeros except in the last column. 2. The vector b is not contained in the span of the columns of A. 7.3 Problems 1. Determine whether or not the following systems of equations have a unique solution, an in…nite number of solutions, or no solution. (a) 3x + 6y = 4 2x 5z = 8 x y z = 10 CHAPTER 7. SYSTEMS OF EQUATIONS 93 (b) 4x y + 8z = 160 17x 8y + 10z = 200 3x + 2y + 2z = 40 (c) 2x 3y = 6 3x + 5z = 15 2x + 6y + 10z = 18 (d) 4x y + 8z = 30 3x + 2z = 20 5x + y 2z = 40 (e) 6x y z = 3 5x + 2y 2z = 10 y 2z = 4 2. Find the values of a for which the following matrices do not have an inverse. (a) 6 2 (b) (c) 1 a 1 5 a 0 @ 4 2 1 A 1 3 1 0 5 3 3 a CHAPTER 7. SYSTEMS OF EQUATIONS (d) 1 1 3 1 @ 0 5 a A 6 2 1 0 94 CHAPTER 8 Using linear algebra in economics 8.1 IS-LM analysis Consider the following model of a closed macroeconomy: Y C I M = = = = C +I +G c((1 t)Y ) i(R) P m(Y; R) with 0 < c0 < 1 1 t i0 < 0 mY > 0; mR < 0 Here Y is GDP, which you can think of as production, income, or spending. The variables C, I, and G are the spending components of GDP, with C 95 CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 96 standing for consumption, I for investment (which is spending by businesses on new capital and by consumers on new housing, not what you do in the stock market), and G for government spending. The …rst equation says that total spending Y is equal to the sum of the components of spending, C + I + G. There are no imports or exports, so this is a closed-economy model. The amount of consumption depends on consumers’ after-tax income, and when Y is income and t is the tax rate, after-tax (or disposable) income is (1 t)Y . So the second equation says that consumption is a function c( ) of disposable income, and c0 > 0 means that it is an increasing function. Investment is typically spending on large items, and it is often …nanced through borrowing. Because of this, investment depends on the interest rate R, and when the interest rate increases borrowing becomes more expensive and the amount of investment falls. Consequently the investment function i(R) is decreasing. M is money supply, and the right-hand side of the fourth equation is money demand. P is the price level, and when things become more expensive it takes more money to purchase the same amount of stu¤. When income Y increases people want to buy more stu¤, and they need more money to do it with, so money demand increases when income increases. Also, since money is cash and checking account balances, which tend not to earn interest, and so when interest rates rise people tend to move some of their assets into interestbearing accounts. This means that money demand falls when interest rates rise. The four equations provide a model of the economy, known as the IS-LM model. The …rst three equations describe the IS curve from macro courses and the fourth equation describes the LM curve. The model is useful for describing a closed economy in the short run, that is, before the price level has time to adjust. At this point you should be wondering why we care. The answer is that we want to do some comparative statics analysis. The variables G, t, and M are exogenous policy variables. P is predetermined and assumed constant for the problem. Everything else is endogenous, so everything else is a function of G, t, and M . We are primarily interested in the variables Y and R, and we want to see how they change when the policy variables change. Let’s look for the comparative statics derivatives dY =dG and dR=dG. To …nd them, …rst simplify the system of four equations to a system of two CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 97 equations: Y = c((1 t)Y ) + i(R) + G M = P m(Y; R) Implicitly di¤erentiate with respect to G to get dY = (1 dG dY dR + i0 +1 dG dG dR dY + P mR 0 = P mY dG dG t)c0 Rearrange as dY dR i0 = 1 dG dG dR dY + mR = 0 mY dG dG We can write this in matrix form dY dG 1 (1 (1 t)c0 mY t)c0 dY dG dR dG i0 mR = 1 0 Now use Cramer’s rule to solve for dY =dG and dR=dG: dY = dG 1 1 i0 0 mR (1 t)c0 i0 mY mR = [1 (1 mR : t)c0 ]mR + mY i0 Both the numerator and denominator are negative, so dY =dG > 0. rises when government spending increases. Now for interest rates: 1 dR = dG 1 (1 t)c0 mY (1 t)c0 mY 1 0 i0 mR = [1 (1 GDP mY : t)c0 ]mR + mY i0 The numerator is negative and so is the denominator. Thus, dR=dG > 0. An increase in government spending increases both GDP and interest rates in the short run. It is also possible to …nd the comparative statics derivatives dY =dt, dR=dt, dY =dM , and dR=dM . You should …gure them out yourselves. CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 8.2 98 Econometrics We want to look at the estimation equation y = X (n k)(k 1) (n 1) + e: (8.1) (n 1) The matrix y contains the data on our dependent variable, and the matrix X contains the data on the independent, or explanatory, variables. Each row is an observation, and each column is an explanatory variable. From the equation we see that there are n observations and k explanatory variables. The matrix is a vector of k coe¢cients, one for each of the k explanatory variables. The estimates will not be perfect, and so the matrix e contains error terms, one for each of the n observations. The fundamental problem in econometrics is to use data to estimate the coe¢cients in in order to make the errors e small. The notion of "small," and the one that is consistent with the standard practice in econometrics, is to make the sum of the squared errors as small as possible. 8.2.1 Least squares analysis We want to minimize the sum of the squared errors. Rewrite (8.1) as e=y and note that T e e= X n X e2i : i=1 Then eT e = (y X )T (y X ) T = yT y X T y yT X + T XT X We want to minimize this expression with respect to the parameter vector . But notice that there is a and a T in the expression. Let’s treat these as two separate variables to get two FOCs: XT y + XT X = 0 T T T y X+ X X = 0 CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 99 These are two copies of the same equation, because the …rst is the transpose of the second. So let’s use the …rst one because it has instead of T . Solving for yields XT X = XT y ^ = (X T X) 1 X T y We call ^ the OLS estimator. Note that it is determined entirely by the data, that is, by the independent variable matrix X and the dependent variable vector y. 8.2.2 A lame example Consider a regression with two observations and one independent variable, with the data given in the table below. Observation Dependent Independent number variable y variable X 1 4 3 2 8 4 There is no constant. Our two observations lead to the two equations 4 = 3 + e1 8 = 4 + e2 We want to …nd the value of that minimizes e21 + e22 . Since e1 = 4 3 and e2 = 8 4 , we get e21 + e22 = (4 = 16 = 80 3 )2 + (8 4 )2 24 + 9 2 + 64 64 + 16 88 + 25 2 : Minimize this with respect to . The FOC is 88 + 50 = 0 44 88 = : = 50 25 2 CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 100 x2 e Xβ y = (4,8) X =(3,4) x1 Figure 8.1: Graphing the lame example in column space Now let’s do it with matrices. The two equations can be written 4 8 = 3 4 ( )+ e1 e2 : The OLS estimator is ^ = (X T X) 1 X T y 3 3 4 = 4 44 = (25) 1 (44) = : 25 1 3 4 4 8 We get the same answer. 8.2.3 Graphing in column space We want to graph the previous example in column space. The example is lame for precisely this reason – so I can graph it in two dimensions. The key here is to think about what we are doing when we …nd the value of to minimize e21 + e22 . X is a point, shown by the vector X = (3; 4) in Figure 8.1, and X is the equation of the line through that point. y is CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 101 another point, shown by the vector y = (4; 8) in the …gure. e is the vector connecting some point X ^ on the line X to the point y. e21 + e22 is the square of the length of e, so we want to minimize the length of e, and we do that by …nding the point on the line X that is closest to the point y. Graphically, the closest point is the one that causes the vector e to be at a right angle to the vector X. Two vectors a and b are orthogonal if a b = 0. This means we can …nd our coe¢cients by having the vector e be orthogonal to the vector X. Remembering that e = y X , we get X (y X ^ ) = 0: Now notice that if we write the two vectors a and b as column matrices A and B, we have a b = AT B: Thus we can rewrite the above expression as X T (y X ^ ) XT y XT X ^ XT X ^ ^ = = = = 0 0 XT y (X T X) 1 X T y; which is exactly the answer we got before. 8.2.4 Interpreting some matrices We have the estimated parameters given by ^ = (X T X) 1 X T y: This tells us that the predicted values of y are y^ = X ^ = X(X T X) 1 X T y: The matrix X(X T X) 1 X T is a projection matrix, and it projects the vector y onto the column space of X. CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 102 The residuals vector can be written e = y X^ = (I X(X T X) 1 X T )y: The two matrices X(X T X) 1 X T and (I X(X T X) 1 X T ) have the special property that they are idempotent, that is, they satisfy the property that AA = A. Geometrically it is clear why this happens. Suppose we apply the projection matrix X(X T X) 1 X T to the vector y. That projects y onto the column space of X, so that X(X T X) 1 X T y lies in the column space of X. If we apply the same projection matrix a second time, it doesn’t do anything because y is already in the column space of X. Similarly, the matrix (I X(X T X) 1 X T ) projects y onto the space that is orthogonal to the column space of X. Doing it a second time does nothing, because it is projecting into the same space a second time. 8.3 Stability of dynamic systems In macroeconomics and time series econometrics a common theme is the stability of the economy. In this section I show how stability works and relate stability to matrices. 8.3.1 Stability with a single variable A dynamic system takes the form of yt+1 = ayt : The variable t denotes the period number, so period t + 1 is the period that immediately follows period t. The variable we are interested in is y, and, in particular, we would like to know how y varies over time. The initial period is period 0, and the initial value of yt is y0 , which for the sake of argument we will assume is positive. The process is not very interesting if a = 0, because then y1 = y2 = y3 = ::: = 0, and it’s also not very interesting if a = 1, because then y1 = y2 = ::: = y0 . We want some movement in yt , so let’s assume that a 6= 0 and a 6= 1. CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 103 The single-variable system is pretty straightforward (some might even say lame), and it follows the following process: y1 = ay0 y2 = ay1 = a2 y0 .. . y t = at y 0 .. . The process y0 ; y1 ; ::: is if it eventually converges to a …nite value, or, in mathematical terms, if there exists a value y such that lim yt = y: t!1 If the process is not stable then it explodes, diverging to either +1 or 1. Whether or not the process is stable is determined by the magnitude of the parameter a. To see how, look at lim yt = lim at y0 = y0 lim at : t!1 t!1 t!1 If a > 1 then at ! 1, and the process cannot be stationary unless y0 just happens to be zero, which is unlikely. Similarly, if a < 1 the process also diverges, this time cycling between positive and negative values depending on whether t is positive or negative, respectively (because y0 is positive). On the other hand, if 0 < a < 1, the value at < 1 and thus yt = at y0 < y0 . What’s more, as t ! 1, the quantity at ! 0, and therefore y t ! 0. Thus, the process is stable when a 2 [0; 1). It also turns out to be stable when 1 < a < 0. The reasoning is the same. When a 2 ( 1; 0) we have t a 2 ( 1; 1) and limt!1 at = 0. This reasoning gives us two stability conditions: jaj < 1 lim yt = 0: t!1 CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 8.3.2 104 Stability with two variables All of that was pretty simple. variables: Let’s look at a dynamic system with two yt+1 = ayt + bzt zt+1 = cyt + dzt Now both variables depend on the past values of both variables. When is this system stable? We can write the system in matrix form: yt+1 zt+1 yt zt a b c d = ; (8.2) or, in shorthand notation, yt+1 = Ayt . But this gives us a really complicated system. Sure, we know that yt zt t a b c d = but the matrix a b c d y0 z0 ; t is really complicated. For example, a b c d 3 = a3 + 2bca + bcd b (a2 + ad + d2 + bc) c (a2 + ad + d2 + bc) d3 + 2bcd + abc and a b c d 10 is too big to …t on the page. Things would be easy if the matrix was diagonal, that is, if b = c = 0. Then we would have two separate single-variable dynamic processes yt+1 = ayt zt+1 = dzt 105 CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS and we already know the stability conditions for these. But the matrix is not diagonal. So let’s mess with things to get a system with a diagonal matrix. It will take a while, so just remember the goal when we get there: we want to generate a system that looks like xt+1 = 1 0 0 2 xt (8.3) because then we can treat the stability of the elements of the vector x separately. 8.3.3 Eigenvalues and eigenvectors Begin by remembering that I is the identity matrix. An eigenvalue of the square matrix A is a value such that det(A I) = a b c d = 0. Taking the determinant yields a b c d = ad a d + 2 bc: So, the eigenvalues are the solutions to the quadratic equation 2 (a + d) + (ad bc) = 0. In general quadratic equations have two solutions, call them For example, suppose that the matrix A is A= 3 6 2 1 : The eigenvalues satisfy the equation 2 (3 1) + ( 3 2 2 12) = 0 15 = 0 = 5; 3: 1 and 2. CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 106 These are the eigenvalues of A. Look at the two matrices we get from the formula A I: 3 5 2 1 5 = 2 6 2 1 ( 3) = 6 2 6 2 6 3 ( 3) 6 2 6 and both of these matrices are singular, which is what you get when you make the determinant equal to zero. An eigenvector of A is a vector v such that (A I)v = 0 where is an eigenvalue of A. For our example, the two eigenvectors are the solutions to 0 v11 2 2 = v21 6 6 0 and v12 v22 6 2 6 2 = 0 0 ; where I made the second subscript denote the number of the eigenvector and the …rst subscript denote the element of that eigenvector. These two equations have simple solutions. The …rst equation holds when v11 v21 because 2 6 = 1 1 1 1 = 2 6 0 0 ; and the second equation holds when v12 v22 because 6 2 6 2 = 1 3 1 3 = 0 0 : CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 107 The relationship between eigenvalues and eigenvectors is useful for our task. Recall that if is an eigenvalue and v is an eigenvector then (A Av I)v = 0 Iv = 0 Av = v: In particular, we have A v11 v21 = A v12 v22 = and v11 v21 1 v12 v22 2 : Construct the matrix V so that the two eigenvectors are columns: v11 v12 v21 v22 V = : Then we have AV = A = 1 = = V v11 v21 v12 v22 A v11 v21 2 v11 v12 v21 v22 1 0 0 2 v12 v22 1 0 0 2 : We are almost there. If V has an inverse, V 1 , we can left-multiply both sides by V 1 to get 0 1 V 1 AV = : (8.4) 0 2 This is the diagonal matrix we were looking for to make our dynamic system easy. CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 8.3.4 108 Back to the dynamic system Go back to the dynamic system yt+1 = Ayt : Create a di¤erent vector x according to x=V 1 y, where V is the matrix of eigenvectors we constructed above. This implies that y = V x: (8.5) Then xt+1 = V 1 yt+1 : Since yt+1 = Ayt , we get xt+1 = V 1 (Ayt ) = V 1 A yt = (V 1 A) (V xt ) = V 1 AV xt : 1 We …gured out the formula for V AV in equation (8.4), which gives us xt+1 = 1 0 0 2 xt : But this is exactly the diagonal system we wanted in equation (8.3). So we are there. It’s about time. Let’s look back at what we have. We began with a matrix A. We found its two eigenvalues 1 and 2 , and we found the two corresponding eigenvectors and combined them to form the matrix V . All of this comes from the matrix A, so we haven’t added anything that wasn’t in the problem. But our original problem was about the vector yt , and our new problem is about the vector xt . Using the intuition we gained from the section with a single-variable dynamic system, we say that the process y0 ; y1 ; ::: is stable if lim yt = 0: t!1 CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 109 The dynamic process yt+1 = Ayt was hard to work with, and so it was di¢cult to determine whether or not it was stable. But the dynamic process wt+1 xt+1 = 1 0 0 2 wt xt is easy to work with, because multiplying it out yields the two simple, singlevariable equations wt+1 = xt+1 = 1 wt 2 xt : And we know the stability conditions for single-variable equations. The …rst one is stable if j 1 j < 1, and the second one is stable if j 2 j < 1. So, if these two equations hold, we have wt xt lim xt = lim t!1 t!1 = 0 0 : Finally, remember that we constructed xt according to (see equation (8.5)) yt = V xt ; and so lim yt = lim V xt = V lim xt = 0 t!1 t!1 t!1 and the original system is also stable. That was a lot of steps, but it all boils down to something simple, and it works for more than two dimensions. The dynamic system yt+1 = Ayt is stable if all of the eigenvalues of A have magnitude smaller than one. 8.4 Problems 1. Consider the following IS-LM model: Y C I M = = = = C +I +G c((1 t)Y ) i(R) P m(Y; R) CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 110 with c0 > 0 i0 < 0 mY > 0; mR < 0 The variables G, t, and M are exogenous policy variables. P is predetermined and assumed constant for the problem. (a) Assume that (1 t)c0 < 1, so that a $1 increase in GDP leads to less than a dollar increase in spending. Compute and interpret dY =dt and dR=dt. (b) Compute and interpret dY =dM and dR=dM . (c) Compute and interpret dY =dP and dR=dP . 2. Consider a di¤erent IS-LM model, this time for an open economy, where X is net exports and T is total tax revenue (as opposed to t which was the marginal tax rate). Y C I X M = = = = = C +I +G+X c(Y T ) i(R) x(Y; R) P m(Y; R) with c0 i0 xY mY > < < > 0 0 0; xR < 0 0; mR < 0 The variables G, T , and M are exogenous policy variables. predetermined and assumed constant for the problem. P is (a) For this problem assume that c0 + xY < 1, so that a $1 increase in GDP leads to less than a dollar increase in spending. Compute and interpret dY =dG and dR=dG. CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 111 (b) Compute and interpret dY =dT and dR=dT . (c) Compute and interpret dY =dM and dR=dM . 3. The following is a model of the long-run economy: Y C I X M Y = = = = = = C +I +G+X c((1 t)Y ) i(R) x(Y; R) P m(Y; R) Y with c0 i0 xY mY > < < > 0 0 0; xR < 0 0; mR < 0 The variables G, t, and M are exogenous policy variables, and Y is also exogenous but not a policy variable. It is interpreted as potential GDP, or full-employment GDP. The variables Y; C; I; X; P are all endogenous. (a) Compute and …nd the signs of dY =dG, dR=dG, and dP=dG. (b) Compute and …nd the signs of dY =dM , dR=dM , and dP=dM . 4. Consider the following system of equations: qD = D(p; I) qS = S(p; w) q D = qS The …rst equation says that the quantity demanded in the market depends on the price of the good p and household income I. Consistent with this being a normal good, we have Dp < 0 and DI > 0. The second equation says that the quantity supplied in the market depends on CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 112 the price of the good and the wage rate of the work force, w. We have Sp > 0 and Sw < 0. The third equation says that markets must clear, so that quantity demanded equals quantity supplied. Three variables are endogenous: qD , qS , and p. Two variables are exogenous: I and w. (a) Show that the market price increases when income increases. (b) Show that the market price increases when the wage rate increases. 5. Find the coe¢cients for a regression based on the following data table: Observation number x1 1 1 2 1 3 1 x2 9 4 3 y 6 2 5 6. Consider a regression based on the following data table: Observation number x1 1 2 2 6 3 -4 x2 y 8 1 24 0 -16 -1 (a) Show that the matrix X T X is not invertible. (b) Explain intuitively, using the idea of a column space, why there is no unique coe¢cient vector for this regression. 7. Find the coe¢cients for a regression based on the following data table: Observation number x1 1 5 2 2 3 3 x2 4 1 6 y 10 2 7 8. Suppose that you are faced with the following data table: Observation number x1 1 1 2 1 3 1 4 1 x2 2 3 5 4 y 10 0 25 15 CHAPTER 8. USING LINEAR ALGEBRA IN ECONOMICS 113 You are thinking about adding one more explanatory variable, x3 , to the regression. x3 is given by 0 1 24 B 36 C C x3 = B @ 60 A 48 Explain why this would be a bad idea. 9. Find the eigenvalues and eigenvectors of the following matrices: (a) 5 1 4 2 (b) 10 12 1 3 (c) C = 3 2 (d) D = 3 4 0 2 (e) E = 10 3 6 4 4 6 10. Consider the dynamic system yt+1 = Ayt : (a) If A = 1=3 0 0 1=5 , is the dynamic system stable? (b) If A = 4=3 1=3 1=4 1=4 , is the dynamic system stable? (c) If A = 1=4 0 1 2=3 , is the dynamic system stable? (d) If A = 1=5 1 2 8=9 , is the dynamic system stable? CHAPTER 9 Second-order conditions 9.1 Taylor approximations for R ! R Consider a di¤erentiable function f : R ! R. Since f is di¤erentiable at point x0 it has a linear approximation at that point. Note that this is a linear approximation to the function, as opposed to a linear approximation at the point. The linear approximation can be written L(x) = f (x0 ) + f 0 (x0 )(x x0 ). When x = x0 we get the original point, so L(x0 ) = f (x0 ). When x is not equal to x0 we multiply the di¤erence (x x0 ) by the slope at point x0 , then add that amount to f (x0 ). All of this is shown in Figure 9.1. The linear approximation L(x) is a straight line that is tangent to the function f (x) at f (x0 ). To get the approximation at point x, take the horizontal distance (x x0 ) and multiply it by the slope of the tangent line, which is just f 0 (x0 ). This gives us the quantity f 0 (x0 )(x x0 ), which must be added to f (x0 ) to get the right linear approximation. 114 115 CHAPTER 9. SECOND-ORDER CONDITIONS f(x) L(x) L(x) f(x) f’(x0)(x – x0) f(x0) x x0 x x – x0 Figure 9.1: A …rst-order Taylor approximation The Taylor approximation builds on this intuition. Suppose that f is n times di¤erentiable. Then an n-th order approximation is f (x) f (x0 ) + f 0 (x0 ) (x 1! x0 ) + f 00 (x0 ) (x 2! x0 )2 + ::: + f (n) (x0 ) (x n! x0 ) n . We care mostly about second-degree approximations, or f (x) f (x0 ) + f 0 (x0 )(x x0 ) + f 00 (x0 ) (x 2 x0 ) 2 : The key to understanding the numbers in the denominators of the different terms is noticing that the …rst derivative of the linear approximation matches the …rst derivative of the function, the second derivative of the second-order approximation equals the second derivative of the original function, and so on. So, for example, we can write the third-degree approximation of f (x) at x = x0 as 0 g(x) = f (x0 ) + f (x0 )(x f 00 (x0 ) x0 ) + (x 2 f 000 (x0 ) x0 ) + (x 6 2 x0 ) 3 : 116 CHAPTER 9. SECOND-ORDER CONDITIONS Di¤erentiate with respect to x to get f 000 (x0 ) g (x) = f (x0 ) + f (x0 )(x x0 ) + (x 2 g 00 (x) = f 00 (x0 ) + f 000 (x0 )(x x0 ) g 000 (x) = f 000 (x0 ) 0 9.2 0 00 x0 ) 2 Second order conditions for R ! R Suppose that f (x) is maximized when x = x0 . Take a second-order Taylor approximation: f (x) f (x0 ) + f 0 (x0 )(x x0 ) + f 00 (x0 ) (x 2 x0 ) 2 : Since f is maximized when x = x0 , the …rst-order condition has f 0 (x0 ) = 0. Thus the second term in the Taylor approximation disappears. We are left with f 00 (x0 ) f (x) f (x0 ) + (x x0 )2 : 2 If f is maximized, it must mean that any departure from x0 leads to a decrease in f . In other words, f 00 (x0 ) (x 2 for all x. Simplifying gives us f (x0 ) + f 00 (x0 ) (x 2 x0 ) 2 x0 ) 2 f (x0 ) 0 and, since (x x0 )2 0, it must be the case that f 00 (x0 ) 0. This is how we can get the second order condition from the Taylor approximation. 9.3 Taylor approximations for Rm ! R This time we are only going to look for a second-degree approximation. We need some notation: 0 1 f1 (x) B C .. rf (x) = @ A . fm (x) 117 CHAPTER 9. SECOND-ORDER CONDITIONS and is called the gradient of f at x. Also 1 0 f11 (x) f1m (x) C B .. .. .. H(x) = @ A . . . fm1 (x) fmm (x) is called the Hessian. It is the matrix of second derivatives. A second-order Taylor approximation for a function of m variables can be written f (x) f (x0 ) + m X m fi (x0 )(xi x0i ) + i=1 m 1 XX fij (x0 )(xi 2 i=1 j=1 x0i )(xj Let’s write this in matrix notation. The …rst term is simply f (x0 ). second term is (x x0 ) rf (x0 ) = (x x0 )T rf (x0 ): x0j ) The The third term is 1 (x x0 )T H(x0 )(x x0 ): 2 Let’s check to make sure this last one works. First check the dimensions, which are (1 m)(m m)(m 1) = (1 1), which is what we want. Then break the problem down. (x x0 )T H(x0 ) is a (1 m) matrix with element j given by m X fij (x0 )(xi x0i ): i=1 0 T 0 To get (x x ) H(x )(x x0 ) we multiply each element of (x by the corresponding element of (x x0 ) and sum, to get m X m X fij (x0 )(xi x0i )(xj x0 )T H(x0 ) x0j ): i=1 j=1 So, the second degree Taylor approximation of the function f : Rm ! R is given by f (x) f (x0 ) + (x 1 x0 )T rf (x0 ) + (x 2 x0 )T H(x0 )(x x0 ) 118 CHAPTER 9. SECOND-ORDER CONDITIONS 9.4 Second order conditions for Rm ! R Suppose that f : Rm ! R is maximized when x = x0 . Then the …rst-order condition is rf (x0 ) = 0 and the second term in the Taylor approximation drops out. maximized when x = x0 it must be the case that 1 f (x0 ) + (x x0 )T H(x0 )(x x0 ) f (x0 ) 2 or (x x0 )T H(x0 )(x x0 ) 0. 9.5 For f to be Negative semide…nite matrices The matrix A is negative semide…nite if, for every column matrix x, we have xT Ax 0. Obviously, the second order condition for a maximum is that H(x0 ) is negative semide…nite. In the form it is written in, though, it is a di¢cult thing to check. Form a submatrix Ai from the square matrix A by keeping the square matrix formed by the …rst i rows and …rst i columns. (Note that this is di¤erent from the submatrix we used to …nd determinants in Section 6.3.) The determinant of Ai is called the i-th leading principal minor of A. Theorem 11 Let A be a symmetric m m matrix. Then A is negative semide…nite if and only if its m leading principle minors alternate in sign so that ( 1)i jAi j 0 for i = 1; :::; m. There are other corresponding notions: A is negative de…nite if xT Ax < 0 for all nonzero vectors x, and this occurs if and only if its m leading principle minors alternate in sign so that ( 1)i jAi j > 0 for i = 1; :::; m. 119 CHAPTER 9. SECOND-ORDER CONDITIONS A is positive de…nite if xT Ax > 0 for all nonzero vectors x, and this occurs if and only if its m leading principle minors are positive, so that jAi j > 0 for i = 1; :::; m. A is positive semide…nite if xT Ax 0 for all vectors x, and this occurs if and only if its m leading principle minors are nonnegative, so that jAi j 0 for i = 1; :::; m. A is inde…nite if none of the other conditions holds. 9.5.1 Application to second-order conditions Suppose that the problem is to choose x1 and x2 to maximize f (x1 ; x2 ). The FOCs are f1 = 0 f2 = 0 and the SOC is H= f11 f21 f21 f22 is negative semide…nite. Note that f21 appears in both o¤-diagonal elements, which is okay because f12 = f21 . That is, it doesn’t matter if you di¤erentiate f …rst with respect to x1 and then with respect to x2 or the other way around. The requirements for H to be negative semide…nite are f11 f11 f12 f21 f22 0 = f11 f22 2 f12 Note that the two conditions together imply that f22 0 0. CHAPTER 9. SECOND-ORDER CONDITIONS 9.5.2 120 Examples 3 3 is negative de…nite because a11 < 0 and a11 a22 3 4 a12 a21 = 3 > 0. 6 1 is positive de…nite because a11 > 0 and a11 a22 a12 a21 = A= 1 3 17 > 0. 5 3 A = is inde…nite because a11 < 0 and a11 a22 a12 a21 = 3 4 29 < 0. A = 9.6 Concave and convex functions All of the second-order conditions considered so far rely on the objective function being twice di¤erentiable. In the single-variable case we require that the second derivative is nonpositive for a maximum and nonnegative for a minimum. In the many-variable case we require that the matrix of second and cross partials (the Hessian) is negative semide…nite for a maximum and positive semide…nite for a minimum. But objective functions are not always di¤erentiable, and we would like to have some second order conditions that work for these cases, too. Figure 9.2 shows a function with a maximum. It also has the following property. If you choose any two points on the curve, such as points a and b, and draw the line segment connecting them, that line segment always lies below the curve. When this property holds for every pair of points on the curve, we say that the function is concave. It is also possible to characterize a concave function mathematically. Point a in the …gure has coordinates (xa ; f (xa )), and point b has coordinates (xb ; f (xb )). Any value x between xa and xb can be written as x = txa + (1 t)xb for some value t 2 [0; 1]. Such a point is called convex combination of xa and xb . When t = 1 we get x = 1 xa + 0 xb = xa , and when t = 0 we get x = 0 xa + 1 xb = xb . When t = 21 we get x = 12 xa + 21 xb which is the midpoint between xa and xb , as shown in Figure 9.2. The points on the line segment connecting a and b in the …gure have 121 CHAPTER 9. SECOND-ORDER CONDITIONS f(x) b f(½xa + ½xb) ½f(xa) + ½f(xb) f(x) a x xa ½xa + ½xb xb Figure 9.2: A concave function coordinates (txa + (1 t)xb ; tf (xa ) + (1 t)f (xb )) for t 2 [0; 1]. Let’s choose one value of t, say t = 12 . The point on the line segment is 1 1 1 1 xa + xb ; f (xa ) + f (xb ) 2 2 2 2 and it is the midpoint between points a and b. But 21 xa + 21 xb is just a value of x, and we can evaluate f (x) at x = 12 xa + 21 xb . According to the …gure, 1 1 f ( xa + xb ) 2 2 1 1 f (xa ) + f (xb ) 2 2 where the left-hand side is the height of the curve and the right-hand side is the height of the line segment connecting a to b. Concavity says that this is true for all possible values of t, not just t = 12 . De…nition 2 A function f (x) is concave if, for all xa and xb , f (txa + (1 for all t 2 [0; 1]. t)xb ) tf (xa ) + (1 t)f (xb ) 122 CHAPTER 9. SECOND-ORDER CONDITIONS f(x) f(x) a b x Figure 9.3: A convex function Concave functions tend to be upward sloping, downward sloping, or have a maximum (that is, upward sloping and then downward sloping). Thus, instead of assuming that a function has the right properties of its second derivative, we can instead assume that it is concave. And notice that nothing in the de…nition says anything about whether x is single-dimensional or multidimensional. The same de…nition of concave works for both singlevariable and multi-variable optimization problems. If a concave function is twice di¤erentiable, it has a nonpositive second derivative. A convex function has the opposite property: the line segment connecting any two points on the curve lies above the curve, as in Figure 9.3. We get a corresponding de…nition. De…nition 3 A function f (x) is convex if, for all xa and xb , f (txa + (1 t)xb ) tf (xa ) + (1 t)f (xb ) for all t 2 [0; 1]. Convexity is the appropriate assumption for minimization, and if a convex function is twice di¤erentiable its second derivative is nonnegative. As an example, consider the standard pro…t-maximization problem, where output is q and pro…t is given by (q) = r(q) c(q); 123 CHAPTER 9. SECOND-ORDER CONDITIONS $ c(q) r(q) q Figure 9.4: Pro…t maximization with a concave revenue function and a convex cost function where r(q) is the revenue function and c(q) is the cost function. The standard assumptions are that the revenue function is concave and the cost function is convex. If both functions are twice di¤erentiable, the …rst-order condition is the familiar r0 (q) c0 (q) = 0 and the second-order condition is r00 (q) c00 (q) 0: This last expression holds if r00 (q) 0 which occurs when r(q) is concave, and if c00 (q) 0 which occurs when c(q) is convex. Figure 9.4 shows the standard revenue and cost functions in a pro…t maximization problem, and in the graph the revenue function is concave and the cost function is convex. Some functions are neither convex nor concave. More precisely, they have some convex portions and some concave portions. Figure 9.5 provides an example. The function is convex to the left of x0 and concave to the right of x0 . 124 CHAPTER 9. SECOND-ORDER CONDITIONS f(x) f(x) x x0 Figure 9.5: A function that is neither concave nor convex 9.7 Quasiconcave and quasiconvex functions The important property for a function with a maximum, such as the one shown in Figure 9.2, is that it rise and then fall. But the function depicted in Figure 9.6 also rises then falls, and clearly has a maximum. But it is not concave. Instead it is quasiconcave, which is the weakest second-order condition we can use. The purpose of this section is to de…ne the terms "quasiconcave" and "quasiconvex," which will take some work. Before we can de…ne them, we need to de…ne a di¤erent term. A set S is convex if, for any two points x; y 2 S , the point x + (1 )y 2 S for all 2 [0; 1]. Let’s break this into pieces using Figure 9.7. In the left-hand graph, the set S is the interior of the oval, and choose two points x and y in S. These can be either in the interior of S or on its boundary, but the ones depicted are in the interior. The set fzjz = x + (1 )y for some 2 [0; 1]g is just the line segment connecting x to y, as shown in the …gure. The set S is convex if the line segment is inside of S, no matter which x and y we choose. Or, using di¤erent terminology, the set S is convex if any convex combination of two points in S is also in S. In contrast, the set S in the right-hand graph in Figure 9.7 is not convex. Even though points x and y are inside of S, the line segment connecting them passes outside of S. In this case the set is nonconvex (there is no such thing as a concave set). 125 CHAPTER 9. SECOND-ORDER CONDITIONS f(x) f(x) x Figure 9.6: A quasiconcave function f(x) f(x) y y S x S x x x Figure 9.7: Convex sets: The set in the left-hand graph is convex because all line segments connecting two points in the set also lie completely within the set. The set in the right-hand graph is not convex because the line segment drawn does not lie completely within the set. 126 CHAPTER 9. SECOND-ORDER CONDITIONS f(x) f(x) y x x1 x2 B(y) Figure 9.8: De…ning quasiconcavity: For any value y, the set B(y) is convex The graphs in Figure 9.7 are for 2-dimensional sets. The de…nition of convex, though, works in any number of dimensions. In particular, it works for 1-dimensional sets. A 1-dimensional set is a subset of the real line, and it is convex if it is an interval, either open, closed, or half-open/half-closed. Now look at Figure 9.8, which has the same function as in Figure 9.6. Choose any value y, and look for the set B(y) = fxjf (x) yg: This is a better-than set, and it contains the values of x that generate a value of f (x) that is at least as high as y. In Figure 9.8 the set B(y) is the closed interval [x1 ; x2 ], which is a convex set. This gives us our de…nition of a quasiconcave function. De…nition 4 The function f (x) is quasiconcave if for any value y, the set B(y) = fxjf (x) yg is convex. To see why quasiconcavity is both important and useful, ‡ip way back to the beginning of the book to look at Figure 1.1 on page 2. That graph depicted either a consumer choice problem, in which case the line is a budget line and the curve is an indi¤erence curve, or it depicted a …rm’s costminimization problem, in which case the line is an isocost line and the curve 127 CHAPTER 9. SECOND-ORDER CONDITIONS f(x) f(x) y x x1 x2 W(y) Figure 9.9: A quasiconvex function is an isoquant. Think about it as a consumer choice problem. The area above the indi¤erence curve is the set of points the consumer prefers to those on the indi¤erence curve. So the set of points above the indi¤erence curve is the better-than set. And it’s convex. So the appropriate second-order condition for utility maximization problems is that the utility function is quasiconcave. Similarly, an appropriate second-order condition for costminimization is that the production function (the function that gives you the isoquant) is quasiconcave. When you take microeconomics, see where quasiconcavity shows up as an assumption. Functions can also be quasiconvex. De…nition 5 The function f (x) is quasiconvex if for any value y, the set W (y) = fxjf (x) yg is convex. Quasiconvex functions are based on worse-than sets W (y). This time the set of points generating values lower than y must be convex. To see why, look at Figure 9.9. This time the points that generate values of f (x) lower than y form an interval, but the better-than set is not an interval. Concave functions are also quasiconcave, which you can see by looking at Figure 9.2, and convex functions are also quasiconvex, which you can see by looking at Figure 9.3. But a quasiconcave function may not be concave, as 128 CHAPTER 9. SECOND-ORDER CONDITIONS in Figure 9.8, and a quasiconvex function may not be convex, as in Figure 9.9. The easiest way to remember the de…nition for quasiconcave is to draw a concave function with a maximum. We know that it is also quasiconcave. Choose a value of y and draw the horizontal line, like we did in Figure 9.8. Which set is convex, the better-than set or the worse-than set? As shown in the …gure, it’s the better-than set that is convex, so we get the right de…nition. If you draw a convex function with a minimum and follow the same steps you can …gure out the right de…nition for a quasiconvex function. 9.8 Problems 1. Find the gradient of f (x1 ; x2 ; x3 ) = 2x1 x23 + 3x1 x22 4x21 and then evaluate it at the point (5; 2; 0). 2. Find the second-degree Taylor approximations of the following functions at x0 = 1: (a) f (x) = 2x3 (b) f (x) = 10x 5x + 9 p 40 x + ln x (c) f (x) = ex 3. Find the second-degree Taylor approximation of the function f (x) = 3x3 4x2 2x + 12 at x0 = 0. 4. Find the second-degree Taylor approximation of the function f (x) = ax2 + bx + c at x0 = 0. 5. Tell whether the following matrices are negative de…nite, negative semide…nite, positive semide…nite, positive de…nite, or inde…nite. (a) (b) 3 2 2 1 1 2 2 4 CHAPTER 9. SECOND-ORDER CONDITIONS 3 4 4 3 4 (d) @ 0 1 0 3 2 (c) 0 1 1 2 A 1 (e) 6 1 1 3 (f) 4 16 16 4 2 1 (g) 129 1 4 0 3 2 @ 2 4 (h) 3 0 1 3 0 A 1 6. State whether the second-order condition is satsi…ed for the following problems. (a) minx;y 4y 2 xy (b) maxx;y 7 + 8x + 6y (c) maxx;y 5xy x2 y2 2y 2 (d) minx;y 6x2 + 3y 2 7. Is (6; 2) a convex combination of (11; 4) and ( 1; 0)? Explain. 8. Use the formula for convexity, and not the second derivative, to show that the function f (x) = x2 is convex. PART III ECONOMETRICS (probability and statistics) CHAPTER 10 Probability 10.1 Some de…nitions An experiment is an activity that involves doing something or observing something resulting in an outcome. The performance of an experiment is a trial. Experiments can be physical, biological, social, or anything else. The sample space for an experiment is the set of possible outcomes of the experiment. The sample space is denoted (the Greek letter omega) and a typical element, or outcome, is denoted ! (lower case omega). The impossible event is the empty set, ?. Suppose, for example, that the experiment consists of tossing a coin twice. The sample space is = f(H; H); (H; T ); (T; H); (T; T )g and each pair is an outcome. Another experiment is an exam. Scores are out of 100, so the sample space is = f0; 1; :::; 100g: 131 CHAPTER 10. PROBABILITY 132 We are going to work with subsets, and we are going to work with some mathematical symbols pertaining to subsets. The notation is given in the following table. In math !2A A\B A[B A B A B AC In English omega is an element of A A intersection B A union B A is a strict subset of B A is a weak subset of B The complement of A An event is a subset of the sample space. If the experiment consists of tossing a coin twice, the event that there is at least one head can be written A = f(H; H); (H; T ); (T; H)g. Note that the entire sample space is an event, and so is the impossible event ?. Sometimes we want to talk about single-element events, and write ! instead of f!g. It is best to think of outcomes and events as occurring. For the latter, an event A occurs if there is some outcome ! 2 A such that ! occurs. Our eventual goal is to assign probabilities to events. To do this we need notation for the set of all possible events. Call it (for sigma-algebra, which is a concept we will not get into). Two events A and B are mutually exclusive if there is no outcome that is in both events, that is, A \ B = ?. If A and B are mutually exclusive then if event A occurs event B is impossible, and vice versa. 10.2 De…ning probability abstractly A probability measure is a mapping P : ! [0; 1], that is, a function mapping events into numbers between 0 and 1. The function P has three key properties: Axiom 1 P (A) 0 for any event A 2 . Axiom 2 P ( ) = 1 133 CHAPTER 10. PROBABILITY Axiom 3 If A1 ; A2 ; ::: is a (possibly …nite) sequence of mutually exclusive events, then P (A1 [ A2 [ :::) = P (A1 ) + P (A2 ) + ::: The …rst axiom states that probabilities cannot be negative. The second one states that the probability that something happens is one. The third axiom states that when events are mutually exclusive, the probability of the union is simply the sum of the probabilities. These three axioms imply some other properties. Theorem 12 P ( ?) = 0: Proof. The events and ? are mutually exclusive since [ ? = , axiom 3 implies that \ ? = ?. Since 1 = P ( [ ?) = P ( ) + P (?) = 1 + P (?), and it follows that P ( ?) = 0. The next result concerns the relation A is contained in B or A is equal to B. Theorem 13 If A B then P (A) , where A B means that either P (B). Proof. Suppose that A and B are events with A B. De…ne C = f! : ! 2 B but ! 2 = Ag. Then A and C are mutually exclusive with A [ C = B, and axiom 3 implies that P (B) = P (A [ C) = P (A) + P (C) P (A). For the next theorems, let AC denote the complement of the event A, that is, AC = f! 2 : ! 2 = Ag. Theorem 14 P (AC ) = 1 P (A). 134 CHAPTER 10. PROBABILITY Proof. Note that AC [ A = and AC \ A = ?. Then P (AC [ A) = P (AC ) + P (A) = 1, and therefore P (AC ) = 1 P (A). Note that this theorem implies that P (A) 1 for any event A. To see why, …rst write P (A) = 1 P (AC ), and by axiom 1 we have P (AC ) 0. The result follows. Theorem 15 P (A [ B) = P (A) + P (B) P (A \ B). Proof. First note that A [ B = A [ (AC \ B). You can see this in Figure 10.1. The points in A [ B are either in A or they are outside of A but in B. The events A and AC \ B are mutually exclusive. Axiom 3 says P (A [ B) = P (A) + P (AC \ B): Next note that B = (A\B)[(AC \B). Once again this is clear in Figure 10.1, but it says that we can separate B into two parts, the part that intersects A and the part that does not. These two parts are mutually exclusive, so P (B) = P (A \ B) + P (AC \ B). (10.1) Rearranging yields P (AC \ B) = P (B) P (A \ B): Substituting this into equation (10.1) yields P (A [ B) = P (A) + P (AC \ B) = P (A) + P (B) P (A \ B): 10.3 De…ning probabilities concretely The previous section told us what the abstract concept probability measure means. Sometimes we want to know actual probabilities. How do we get them? The answer relies on the ability to partition the sample space into equally likely outcomes. 135 CHAPTER 10. PROBABILITY A ∩ BC A∩B A AC ∩ B B Ω Figure 10.1: Finding the probability of the union of two sets Theorem 16 Suppose that every outcome in the sample space = f! 1 ; :::; ! n g is equally likely. Then the probability of any event A is the number of outcomes in A divided by n. Proof. We know that P ( ) = P (f! 1 g [ [ f! n g) = 1. By construction f! i g \ f! j g = ? when i 6= j, and so the events f! 1 g; :::; f! n g are mutually exclusive. By Axiom 3 we have P ( ) = P (f! 1 g) + ::: + P (f! n )g = 1: Since each of the outcomes is equally likely, this implies that P (f! i g) = 1=n for i = 1; :::; n. If event A contains k of the outcomes in the set f! 1 ; :::; ! n g, it follows that P (A) = k=n. This theorem allows us to compute probabilities from experiments like ‡ipping coins, rolling dice, and so on. For example, if a die has six sides, the probability of the outcome 5 is 1=6. The probability of the event f1; 2g is 1=6 + 1=6 = 1=3, and so on. For an exercise, …nd the probability of getting exactly one head in four tosses of a fair coin. Answer: There are 16 possible outcomes. Four of them have one head. So, the probability is 1=4. 136 CHAPTER 10. PROBABILITY In general events are not equally likely, so we cannot determine probabilities theoretically in this manner. Instead, the probabilities of the events are given directly. 10.4 Conditional probability Suppose that P (B) > 0, so that the event B occurs with positive probability. Then P (AjB) is the conditional probability that event A occurs given that event B has occurred. It is given by the formula P (AjB) = P (A \ B) : P (B) Think about what this expression means. The numerator is the probability that both A and B occur. The denominator is the probability that B occurs. Clearly A \ B is a subset of B, so the numerator is smaller than the denominator, as required. The ratio can be interpreted as the fraction of the time when B occurs that A also occurs. Consider the probability distribution given in the table below. Outcomes are two dimensional, based on the values of x and y. The probabilities of the di¤erent outcomes are given in the entries of the table. Note that all the entries are nonnegative and that the sum of all the entries is one, as required for a probability measure. x=1 x=2 x=3 x=4 y=1 0.02 0.05 0.04 0.10 y=2 0.01 0.00 0.15 0.16 y=3 0.02 0.03 0.02 0.02 y=4 0.10 0.11 0.09 0.08 Find the conditional probability P (x = 2jy = 4). The formula is P (x = 2 and y = 4)=P (y = 4). The probability that x = 2 and y = 4 is just the entry in a single cell, and is 0.11. The probability that y = 4 is the sum of the probabilities in the last column, or 0.38. So, P (x = 2jy = 4) = 0:11=0:38 = 11=38. Now …nd the conditional probability that y is either 1 or 2 given that x 3. The probability that y 2 and x 3 is the sum of the four cells 137 CHAPTER 10. PROBABILITY in the lower left, or 0.45. The probability that x 3 is 0.66. So, the conditional probability is 45=66 = 15=22. Now look at a medical example. A patient can have condition A or not. He takes a test which turns out positive or not. The probabilities are given in the following table: Test positive Test negative Condition A 0.010 0.002 0.001 0.987 Healthy Note that condition A is quite rare, with only 12 people in 1000 having it. Also, a positive test is ten times more likely to come from a patient with the condition than from a patient without the condition. We get the following conditional probabilities: P (Ajpositive) P (healthyjnegative) P (positivejA) P (negativejhealthy) 10.5 = = = = 10=11 = 0:909; 987=989 = 0:998; 10=12 = 0:833; 987=988 = 0:999: Bayes’ rule Theorem 17 (Bayes’ rule) Assume that P (B) > 0. Then P (AjB) = P (BjA)P (A) : P (B) Proof. Note that P (AjB) = P (A \ B) P (B) P (BjA) = P (A \ B) : P (A) and Rearranging the second one yields P (A \ B) = P (BjA)P (A) 138 CHAPTER 10. PROBABILITY and the theorem follows from substitution. Let’s make sure Bayes’ rule works for the medical example given above. We have P (positivejA) = 10=12, P (positive) = 11=1000, and P (A) = 12=1000. Using Bayes’ rule we get P (Ajpositive) = P (positivejA) P (A) = P (positive) 10 12 12 1000 11 1000 = 10 : 11 The interpretation of Bayes’ rule is for responding to updated information. We are interested in the occurrence of event A after we receive some new information. We start with the prior P (A). Then we …nd out that B holds. We should use this new information to update the probability of A occurring. P (AjB) is called the posterior probability. Bayes’ rule tells us how to do this. We multiply the prior P (A) by the likelihood P (BjA) : P (B) If this ratio is greater than one, the posterior probability is higher than the prior probability. If the ratio is smaller than one, the posterior probability is lower. The ratio is greater than one if A occurring makes B more likely. Bayes’ rule is important in game theory, …nance, and macro. People don’t seem to follow it. Here is a famous example (Kahneman and Tversky, 1973 Psychological Review). Some subjects are told that a group consists of 70 lawyers and 30 engineers. The rest of the subjects are told that the group has 30 lawyers and 70 engineers. All subjects were then given the following description: Dick is a 30 year old man. He is married with no children. A man of high ability and high motivation, he promises to be quite successful in his …eld. He is well liked by his colleagues. Subjects were then asked to judge the probability that Dick is an engineer. Subjects in both groups said that it is about 0.5, ignoring the prior information. The new information is uninformative, so P (BjA)=P (B) = 1, and according to Bayes’ rule the posterior should be the same as the prior. This example has people overweighting the new information. Psychologists have also come up with studies in which subjects overweight the prior. When subjects overweight the new information it is called representativeness, and when they overweight the prior it is called conservatism. 139 CHAPTER 10. PROBABILITY 10.6 Monty Hall problem At the end of the game show Let’s Make a Deal the host, Monty Hall, o¤ers a contestant the choice among three doors, labeled A, B, and C. There is a prize behind one of the doors, and nothing behind the other two. After the contestant chooses a door, to build suspense Monty Hall reveals one of the doors with no prize. He then asks the contestant if she would like to stay with her original door or switch to the other one. What should she do? The answer is that she should take the other door. To see why, suppose she chooses door A, and that Monty reveals door B. What is the probability that the prize is behind door C given that door B was revealed? Bayes’ rule says we use the formula P (prize behind Cj reveal B) = P (reveal Bj prize C) P (prize C) : P (reveal B) Before revealing the door, the prize was equally likely to be behind each of the three doors, so P (prize A) = P (prize B) = P (prize C) = 1=3. Next …nd the conditional probability that Monty reveals door B given that the prize is behind door C. Remember that Monty cannot reveal the door with the prize behind it or the door chosen by the contestant. Therefore Monty must reveal door B if the prize is behind door C, and the conditional probability P (reveal Bj prize C) = 1. The remaining piece of the formula is the probability that he reveals B. We can write this as P (reveal B) = P (reveal Bj prize A) P (A) +P (reveal Bj prize B) P (B) +P (reveal Bj prize C) P (C): The middle term is zero because he cannot reveal the door with the prize behind it. The last term is 1=3 for the reasons given above. If the prize is behind A he can reveal either B or C and, assuming he does so randomly, the conditional probability P (reveal Bj prize A) = 1=2. Consequently the …rst term is 12 31 = 16 . Using all this information, the probability of revealing door B is P (reveal B) = 16 + 13 = 12 . Plugging this into Bayes’ rule yields P (prize Cj reveal B) = 1 3 1 1 2 2 = : 3 The probability that the prize is behind A given that he revealed door B is 1 P (prize Cj reveal B) = 1=3. The contestant should switch. CHAPTER 10. PROBABILITY 10.7 140 Statistical independence Two events A and B are independent if and only if P (A\B) = P (A) P (B). Consider the following table relating accidents to drinking and driving. Accident No accident Drunk driver 0.03 0.10 0.03 0.84 Sober driver Notice that from this table that half of the accidents come from sober drivers, but there are many more sober drivers than drunk ones. The question is whether accidents are independent of drunkenness. Compute P (drunk \ accident) = 0:03, P (drunk) = 0:13, P (accident) = 0:06, and …nally P (drunk) P (accident) = 0:13 0:06 = 0:0078 6= 0:03: So, accidents and drunk driving are not independent events. This is not surprising, as we would expect drunkenness to be a contributing factor to accidents. Note that P (accidentjdrunk) = 3=13 = 0:23 while P (accidentjsober) = 3=87 = 0:03 4. We can prove an easy theorem about independent events. Theorem 18 If A and B are independent then P (AjB) = P (A). Proof. We have P (AjB) = P (A)P (B) P (A \ B) = = P (A): P (B) P (B) According to this theorem, if accidents and drunkenness were independent events, then P (accidentjdrunk) = P (accident); that is, the probability of getting in an accident when drunk is the same as the overall probability of getting in an accident. 141 CHAPTER 10. PROBABILITY 10.8 Problems 1. Answer the questions using the table below. x=5 x = 20 x = 30 y=1 0:01 0:03 0:11 y=2 0:03 0:05 0:04 y=3 0:17 0:04 0:02 y=4 0:00 0:20 0:07 y=5 0:00 0:12 0:11 (a) What is the most likely outcome? (b) What outcomes are impossible? (c) Find the probability that x = 30. (d) Find the probability that x 2 f5; 20g and 2 (e) Find the probability that y y 4. 2 conditional on x 20. (f) Verify Bayes’ rule for P (y = 4jx = 20). (g) Are the events x 20 and y 2 f1; 4g statistically independent? 2. Answer the questions from the table below: a=1 a=2 a=3 a=4 a=5 b=1 0:02 0:03 0:01 0:00 0:12 b=2 0:02 0:01 0:01 0:05 0:06 b=3 0:21 0:05 0:01 0:00 0:00 b=4 0:02 0:06 0:06 0:12 0:14 (a) Which event is more likely, A = f(a; b) : 3 f(a; b) : b 2 and a = 5g? a 4g or B = (b) List the largest impossible event. (c) Find the probability that b 6= 3. (d) Find P (b = 2ja = 5). (e) Find P (a 3jb 2 f1; 4g): (f) Are the events a 2 f1; 3g and b 2 f1; 2; 4g statistically independent? CHAPTER 10. PROBABILITY 142 3. A disease hits 1 in every 20,000 people. A diagnostic test is 95% accurate, that is, the test is positive for 95% of people with the disease, and negative for 95% of the people who do not have the disease. Max just tested positive for the disease. What is the probability he has it? 4. You have data that sorts individuals into occupations and age groups. There are three occupations: doctor, lawyer, and entrpreneur. There are two age categories: below 40 (young) and above 40 (old). You wanted to know the probability that an old person is an entrepreneur. Your grad student misunderstands you, though, and presents you with the following information: 20% 40% 20% 70% of of of of the the the the sample are doctors and 30% are entrepreneurs doctors are young entrepreneurs are young lawyers are young Find the probability that the an old person is an entrepreneur. CHAPTER 11 Random variables 11.1 Random variables A random variable is a variable whose value is a real number, and that number is determined by the outcome of an experiment. For example, the number of heads in ten coin tosses is a random variable, and the Dow-Jones Industrial Average is a random variable. The standard notation for a random variable is to place a tilde over the variable. So x~ is a random variable. The realization of a random variable is based on the outcome of an actual experiment. For example, if I toss a coin ten times and …nd four heads, the realization of the random variable is 4. When the random variable is denoted x~ its realization is denoted x. A random variable is discrete if it can take only discrete values (either a …nite number or a countable in…nity of values). A random variable is continuous if it can take any value in an interval. Random variables have probability measures, and we can use random variables to de…ne events. For example, P (~ x = x) is the probability that the 143 144 CHAPTER 11. RANDOM VARIABLES realization of the random variable x~ is x, and P (~ x 2 [2; 3]) is the probability that the realization of the random variable x~ falls in the interval [2; 3]. The event in the latter example is the event that the realization of x~ falls in the interval [2; 3]. 11.2 Distribution functions The distribution function for the random variable x~ with probability measure P is given by F (x) = P (~ x x): The distribution function F (x) tells the probability that the realization of the random variable is no greater than x. Distribution functions are almost always denoted by capital letters. Theorem 19 Distribution functions are nondecreasing and take values in the interval [0; 1]. Proof. The second part of the statement is obvious. For the …rst part, suppose x < y. Then the event x~ x is contained in the event x~ y, and by Theorem 13 we have F (x) = P (~ x 11.3 x) P (~ x y) = F (y): Density functions If the distribution function F (x) is di¤erentiable, the density function is f (x) = F 0 (x): If the distribution function F (x) is discrete, the density function is P (~ x = x) for each possible value of x. Sometimes distribution functions are neither di¤erentiable nor discrete. This causes headaches that we will not deal with here. Note that it is possible to go from a density function to a distribution function: Z x f (t)dt: F (x) = 1 CHAPTER 11. RANDOM VARIABLES 145 So, the distribution function is the accumulated value of the density function. This leads to some additional common terminology. The distribution function is often called the cumulative density function, or c.d.f. The density function is often called the probability density function, or p.d.f. The support of a distribution F (x) is the smallest closed set containing fxjf (x) 6= 0g, that is, the set of points for which the density is positive. For a discrete distribution this is just the set of outcomes to which the distribution assigns positive probability. For a continuous distribution the support is the smallest closed set containing all of the points that have positive probability. For our purposes there is no real reason for using a closed (as opposed to open) set, but the de…nition given here is mathematically correct. 11.4 Useful distributions 11.4.1 Binomial (or Bernoulli) distribution The binomial distribution arises when the experiment consists of repeated trials with the same two possible outcomes in each trial. The most obvious example is ‡ipping a coin n times. The outcome is a series of heads and tails, and the probability distribution governing the number of heads in the series of n coin tosses is the binomial distribution. To get a general formula, label one possible outcome of the trial a success and the other a failure. These are just labels. In coin tossing, we could count a head as a "success" and a tail as a "failure," or we could do it the other way around. If we are checking lightbulbs to see if they work, we could label a working lightbulb as a "success" and a nonworking bulb as a "failure," or we could do it the other way around. A coauthor (Harold Winter) and I used the binomial distribution to model juror bias, and the two possible outcomes were a juror biased toward conviction and a juror biased toward acquittal. We obviously cared about the total bias of the group. We had to arbitrarily label one form of bias a "success" and the other a "failure." Suppose that the probability of a success is p in any given trial, which means that the probability of a failure is q = 1 p. Also, assume that the trials are statistically independent, so that a success in trial t has no e¤ect on the probability of success in period t + 1. The question is, what is the probability of x successes in n trials? Let’s work this out for the case of two trials. Let the vector (outcome 1, 146 CHAPTER 11. RANDOM VARIABLES outcome 2) denote the event in which outcome 1 is realized in the …rst trial and outcome 2 in the second trial. Using P as the probability measure, we have P (success, success) P (success, failure) P (failure, success) P (failure, failure) = = = = p2 pq pq q2 The probability of two successes in two trials is p2 , the probability of one success in two trials is 2pq, and the probability of two failures is q 2 . With three trials, letting s denote a success and f a failure, we have P (s; s; s) P (s; s; f ) P (s; f; f ) P (f; f; f ) = = = = p3 P (s; f; s) = P (f; s; s) = p2 q P (f; s; f ) = P (f; f; s) = pq 2 q3 Thus the probability of three successes is p3 , the probability of two successes is 3p2 q, the probability of one success is 3pq 2 , and the probability of no successes is q 3 . In general, the rule for x successes in n trials when the probability of success in a single trial is p is b(x; n; p) = n x n p q x x There are two pieces of the formula. The probability of a single, particular con…guration of x successes and n x failures is px q n x . For example, the probability that the …rst x trials are successes and the last n x trials are failures is px q n x . The number of possible con…gurations with x successes and n x failures is nx , which is given by n x = 1 2 ::: n n! = x!(n x)! [1 ::: x][1 ::: (n x)] : Note that the function b(x; n; p) is a density function. The binomial distribution is a discrete distribution, so b(x; n; p) = P (~ x = x), where x~ is the random variable measuring the number of successes in n trials. CHAPTER 11. RANDOM VARIABLES 147 The binomial distribution function is B(x; n; p) = x X b(x; n; p); i=0 so that it is the probability of getting x or fewer successes in n trials. Microsoft Excel, and probably similar programs, make it easy to compute the binomial density and distribution. The formula for b(x; n; p) is =BINOMDIST(x; n; p; 0) and the formula for B(x; n; p) is =BINOMDIST(x; n; p; 1) The last argument in the function just tells the program whether to compute the density or the distribution. Let’s go back to my jury example. Suppose we want to draw a pool of 12 jurors from the population, and that 20% of them are biased toward acquittal, with the rest biased toward conviction. The probability of drawing 2 jurors biased toward acquittal and 10 biased toward conviction is b(2; 12; 0:2) = 0:283 The probability of getting at most two jurors biased toward acquittal is B(2; 12; 0:2) = :558 The probability of getting a jury in which every member is biased toward conviction (and no member is biased toward acquittal) is b(0; 12; 0:2) = 0:069 11.4.2 Uniform distribution The great appeal of the uniform distribution is that it is easy to work with mathematically. It’s density function is given by 8 1 x 2 [a; b] < b a f (x) = if : 0 x2 = [a; b] 148 CHAPTER 11. RANDOM VARIABLES The corresponding distribution function is 8 x<a < 0 x a if a x b F (x) = : b a 1 x>b Okay, so these look complicated. But, if x is in the support interval [a; b], then the density function is the constant function f (x) = 1=(b a) and the distribution function is the linear function F (x) = (x a)=(b a). Graphically, the uniform density is a horizontal line. Intuitively, it spreads the probability evenly (or uniformly) throughout the support [a; b], which is why it has its name. The distribution function is just a line with slope 1=(b a) in the support [a; b]. 11.4.3 Normal (or Gaussian) distribution The normal distribution is the one that gives us the familiar bell-shaped density function, as shown in Figure 11.1. It is also central to statistical analysis, as we will see later in the course. We begin with the standard normal distribution. For now, its density function is 1 f (x) = p e 2 x2 =2 and its support is the entire real line: ( 1; 1). The distribution function is Z x 1 2 e t =2 dt F (x) = p 2 1 and we write it as an integral because there is no simple functional form. Later on we will …nd the mean and standard deviation for di¤erent distributions. The standard normal has mean 0 and standard deviation 1. A more general normal distribution with mean and standard deviation has density function 1 2 2 f (x) = p e (x ) =2 2 and distribution function Z x 1 2 2 F (x) = p e (t ) =2 dt: 2 1 149 CHAPTER 11. RANDOM VARIABLES y 0.4 0.3 0.2 0.1 -5 -4 -3 -2 -1 0 1 2 3 4 5 x Figure 11.1: Density function for the standard normal distribution The e¤ects of changing the mean can be seen in Figure 11.2. The peak of the normal distribution is at the mean, so the standard normal peaks at x = 0, which is the thick curve in the …gure. The thin curve in the …gure has a mean of 2.5. Changing the standard deviation has a di¤erent e¤ect, as shown in Figure 11.3. The thick curve is the standard normal with = 1, and the thin curve has = 2. As you can see, increasing the standard deviation lowers the peak and spreads the density out, moving probability away from the mean and into the tails. 11.4.4 Exponential distribution The density function for the exponential distribution is 1 f (x) = e x= and the distribution function is F (x) = 1 e x= : It is de…ned for x > 0. The exponential distribution is often used for the failure rate of equipment: the probability that a piece of equipment will fail 150 CHAPTER 11. RANDOM VARIABLES y 0.3 0.2 0.1 0 -5 -2.5 0 2.5 5 x Figure 11.2: Changing the mean of the normal density y 0.3 0.2 0.1 0 -5 -2.5 0 2.5 5 x Figure 11.3: Increasing the standard deviation of the normal density 151 CHAPTER 11. RANDOM VARIABLES y 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 1 2 3 4 5 x Figure 11.4: Density function for the lognormal distribution by time x is F (x). Accordingly, f (x) is the probability of a failure at time x. 11.4.5 Lognormal distribution The random variable x~ has the lognormal distribution if the random variable y~ = ln x~ has the normal distribution. The density function is 1 h (ln x)2 =2 i 1 e f (x) = p x 2 and it is de…ned for x 11.4.6 0. It is shown in Figure 11.4. Logistic distribution The logistic distribution function is F (x) = Its density function is f (x) = 1 1+e x : e x : (1 + e x )2 152 CHAPTER 11. RANDOM VARIABLES y 0.25 0.2 0.15 0.1 0.05 -5 -2.5 0 2.5 5 x Figure 11.5: Density function for the logistic distribution It is de…ned over the entire real line and gives the bell-shaped density function shown in Figure 11.5. CHAPTER 12 Integration If you took a calculus class from a mathematician, you probably learned two things about integration: (1) integration is the opposite of di¤erentiation, and (2) integration …nds the area under a curve. Both of these are correct. Unfortunately, in economics we rarely talk about the area under a curve. There are exceptions, of course. We sometimes think about pro…t as the area between the price line and the marginal cost curve, and we sometimes compute consumer surplus as the area between the demand curve and the price line. But this is not the primary reason for using integration in economics. Before we get into the interpretation, we should …rst deal with the mechanics. As already stated, integration is the opposite of di¤erentiation. To make this explicit, suppose that the function F (x) has derivative f (x). The following two statements provide the fundamental relationship between derivatives and integrals: Z b f (x)dx = F (b) F (a); (12.1) a 153 154 CHAPTER 12. INTEGRATION and Z f (x)dx = F (x) + c; (12.2) where c is a constant. The integral in (12.1) is a de…nite integral, and its distinguishing feature is that the integral is taken over a …nite interval. The integral in (12.2) is an inde…nite integral, and it has no endpoints. The reason for the names is that the solution in (12.1) is unique, or de…nite, while the solution in (12.2) is not unique. This occurs because when we integrate the function f (x), all we know is the slope of the function F (x), and we do not know anything about its height. If we choose one function that has slope f (x), call it F (x), and we shift it upward by one unit, its slope is still f (x). The role of the constant c in (12.2), then, is to account for the indeterminacy of the height of the curve when we take an integral. The two equations (12.1) and (12.2) are consistent with each other. To see why, notice that Z Z f (x)dx = 1 f (x)dx; 1 so an inde…nite integral is really just an integral over the entire real line ( 1; 1). Furthermore, Z a b f (x)dx = Z b f (x)dx 1 Z a f (x)dx 1 = [F (b) + c] [F (a) + c] = F (b) F (a): Some integrals are straightforward, but others require some work. From our point of view the most important ones follow, and they can be checked by di¤erentiating the right-hand side. Z xn+1 xn dx = + c for n 6= 1 n+1 Z 1 dx = ln x + c x Z erx erx dx = +c r Z ln xdx = x ln x x + c 155 CHAPTER 12. INTEGRATION There are also two useful rules for more complicated integrals: Z Z af (x)dx = a f (x)dx Z Z Z [f (x) + g(x)] dx = f (x)dx + g(x)dx: The …rst of these says that a constant inside of the integral can be moved outside of the integral. The second one says that the integral of the sum of two functions is the sum of the two integrals. Together they say that integration is a linear operation. 12.1 Interpreting integrals Repeat after me (okay, way after me, because I wrote this in April 2008): Integrals are used for adding. They can also be used to …nd the area under a curve, but in economics the primary use is for addition. To see why, suppose we wanted to do something strange like measure the amount of water ‡owing in a particular river during a year. We have not …gured out how to measure the total volume, but we can measure the ‡ow at any instant in time using our Acme Hydro‡owometerTM . At time t the volume of water ‡owing past a particular point as measured by the Hydro‡owometerTM is h(t). Suppose that we break the year into n intervals of length T =n each, where T is the amount of time in a year. We measure the ‡ow once per time interval, and our measurement times are t1 ; :::; tn . We use the measured ‡ow at time ti to calculate the total ‡ow for period i according to the formula h(ti ) T n where h(ti ) is our measure of the instantaneous ‡ow and T =n is the length of time in the interval. The total estimated volume for the year is then V (n) = tn X t=t1 T h(ti ) : n (12.3) 156 CHAPTER 12. INTEGRATION We can make our estimate of the volume more accurate by taking more measurements of the ‡ow. As we do this n becomes larger and T =n becomes smaller. Now suppose that we could measure the ‡ow at every instant of time. Then T =n ! 0, and if we tried to do the summation in equation (12.3) we would add up a whole bunch of numbers, each of which is multiplied by zero. But the total volume is not zero, so this cannot be the right approach. It’s not. The right approach uses integrals. The idea behind an integral is adding an in…nite number of zeroes together to get something that is not zero. Our correct formula for the volume would be Z T h(t)dt: V = 0 The expression dt takes the place of the expression T =n in the sum, and it is the length of each measurement interval, which is zero. We can use this approach in a variety of economic applications. One major use is for taking expectations of continuous random variables, which is the topic of the next chapter. Before going on, though, there are two useful tricks involving integrals that deserve further attention. 12.2 Integration by parts Integration by parts is a very handy trick that is often used in economics. It is also what separates us from the lower animals. The nice thing about integration by parts is that it is simple to reconstruct. Start with the product rule for derivatives: d [f (x)g(x)] = f 0 (x)g(x) + f (x)g 0 (x): dx Integrate both sides of this with respect to x and over the interval [a; b]: Z a b d [f (x)g(x)]dx = dx Z a b 0 f (x)g(x)dx + Z b f (x)g 0 (x)dx: (12.4) a Note that the left-hand term is just the integral (with respect to x) of a derivative (with respect to x), and combining those two operations leaves 157 CHAPTER 12. INTEGRATION the function unchanged: Z b d [f (x)g(x)]dx = f (x)g(x)jba dx a = f (b)g(b) f (a)g(a): Plugging this into (12.4) yields Z b Z b b 0 f (x)g 0 (x)dx: f (x)g(x)dx + f (x)g(x)ja = a a Now rearrange this to get the rule for integration by parts: Z b Z b b 0 f (x)g 0 (x)dx: f (x)g(x)dx = f (x)g(x)ja (12.5) a a When you took calculus, integration by parts was this mysterious thing. Now you know the secret – it’s just the product rule for derivatives. 12.2.1 Application: Choice between lotteries Here is my favorite use of integration by parts. You may not understand the economics yet, but you will someday. Suppose that an individual is choosing between two lotteries. Lotteries are just probability distributions, and the individual’s objective function is Z b u(x)F 0 (x)dx (12.6) a where a is the lowest possible payo¤ from the lottery, b is the highest possible payo¤, u is a utility function de…ned over amounts of money, and F 0 (x) is the density function corresponding to the probability distribution function F (x). (I use the derivative notation F 0 (x) instead of the density notation f (x) to make the use of integration by parts more transparent.) The individual can choose between lottery F (x) and lottery G(x), and we would like to …nd properties of F (x) and G(x) that guarantee that the individual likes F (x) better. How can we do this? We must start with what we know about the individual. The individual likes money, and u(x) is the utility of money. If she likes money, u(x) must be nondecreasing, so u0 (x) 0. 158 CHAPTER 12. INTEGRATION And that’s all we know about the individual. Now look back at expression (12.6). It is the integral of the product of u(x) and F 0 (x). But the only thing we know about the individual is that u0 (x) 0, and expression (12.6) does not have u0 (x) in it. So let’s integrate by parts. Z b Z b b 0 u0 (x)F (x)dx: u(x)F (x)dx = u(x)F (x)ja a a To simplify this we need to know a little more about probability distribution functions. Since a is the lowest possible payo¤ from the lottery, the distribution function must satisfy F (a) = 0 (this comes from Theorem 19). Since b is the highest possible payo¤ from the lottery, the distribution function must satisfy F (b) = 1. So, the above expression reduces to Z Z b 0 u(x)F (x)dx = u(b) b u0 (x)F (x)dx: (12.7) a a We can go through the same analysis for the other lottery, G(x), and …nd Z Z b 0 u(x)G (x)dx = u(b) b u0 (x)G(x)dx: (12.8) a a The individual chooses the lottery F (x) over the lottery G(x) if Z Z b 0 u(x)F (x)dx b u(x)G0 (x)dx; a a that is, if the lottery F (x) generates a higher value of the objective function than the lottery G(x) does. Or written di¤erently, she chooses F (x) over G(x) if Z b Z b 0 u(x)G0 (x)dx 0: u(x)F (x)dx a a 159 CHAPTER 12. INTEGRATION Subtracting (12.8) from (12.7) yields Z b Z b 0 u(x)G0 (x)dx u(x)F (x)dx a a Z b Z b 0 u0 (x)G(x)dx u (x)F (x)dx u(b) = u(b) a a Z b Z b u0 (x)F (x)dx u0 (x)G(x)dx = a a Z b u0 (x) [G(x) F (x)] dx: = a The di¤erence depends on something we know about: u0 (x), which we know is nonnegative. The individual chooses F (x) over G(x) if the above expression is nonnegative, that is, if Z b u0 (x) [G(x) F (x)] dx 0: a We know that u0 (x) 0. We can guarantee that the product u0 (x) [G(x) F (x)] is nonnegative if the other term, [G(x) F (x)], is also nonnegative. So, we are certain that she will choose F (x) over G(x) if G(x) F (x) 0 for all x. This turns out to be the answer to our question. Any individual who likes money will prefer lottery F (x) to lottery G(x) if G(x) F (x) 0 for all x. There is even a name for this condition – …rst-order stochastic dominance. But the goal here was not to teach you about choice over lotteries. The goal was to show the usefulness of integration by parts. So let’s look back and see exactly what it did for us. The objective function was an integral of the product of two terms, u(x) and F 0 (x). We could not assume anything about u(x), but we could assume something about u0 (x). So we used integration by parts to get an expression that involved the term we knew something about. And that is its beauty. 12.3 Di¤erentiating integrals As you have probably noticed, in economics we di¤erentiate a lot. Sometimes, though, the objective function has an integral, as with expression 160 CHAPTER 12. INTEGRATION (12.6) above. Often we want to di¤erentiate an objective function to …nd an optimum, and when the objective function has an integral we need to know how to di¤erentiate it. There is a rule for doing so, called Leibniz’s rule, named after the 17th-century German mathematician who was one of the two independent inventors of calculus (along with Newton). We want to …nd Z d b(t) f (x; t)dx: dt a(t) Note that we are di¤erentiating with respect to t, and we are integrating with respect to x. Nevertheless, t shows up three times in the expression, once in the upper limit of the integral, b(t), once in the lower limit of the integral, a(t), and once in the integrand, f (x; t). We need to …gure out what to do with these three terms. A picture helps. Look at Figure 12.1. The integral is the area underneath the curve f (x; t) between the endpoints a(t) and b(t). Three things happen when t changes. First, the function f (x; t) shifts, and the graph shows an upward shift, which makes the integral larger because the area under a higher curve is larger. Second, the right endpoint b(t) changes, and the graph shows it getting larger. This again increases the integral because now we are integrating over a larger interval. Third, the left endpoint a(t) changes, and again the graph shows it getting larger. This time, though, it makes the integral smaller because moving the left endpoint rightward shrinks the interval over which we are integrating. Leibniz’s rule accounts for all three of these shifts. Leibniz’s rule says d dt Z b(t) a(t) f (x; t)dx = Z b(t) a(t) @f (x; t) dx + b0 (t)f (b(t); t) @t a0 (t)f (a(t); t): Each of the three terms corresponds to one of the shifts in Figure 12.1. The …rst term accounts for the upward shift of the curve f (x; t). The term @f (x; t)=@t tells how far upward the curve shifts at point x, and the integral Z b(t) a(t) @f (x; t) dx @t tells how much the area changes because of the upward shift in f (x; t). The second term accounts for the movement in the right endpoint, b(t). Using the graph, the amount added to the integral is the area of a rectangle 161 CHAPTER 12. INTEGRATION f(x,t) f(x,t) x a(t) b(t) Figure 12.1: Leibniz’s rule that has height f (b(t); t), that is, f (x; t) evaluated at x = b(t), and width b0 (t), which accounts for how far b(t) moves when t changes. Since area is just length times width, we get b0 (t)f (b(t); t), which is exactly the second term. The third term accounts for the movement in the left endpoint, a(t). Using the graph again, the change in the integral is the area of a rectangle that has height f (a(t); t) and width a0 (t). This time, though, if a(t) increases we are reducing the size of the integral, so we must subtract the area of the rectangle. Consequently, the third term is a0 (t)f (a(t); t). Putting these three terms together gives us Leibniz’s rule, which looks complicated but hopefully makes sense. 12.3.1 Application: Second-price auctions A simple application of Leibniz’s rule comes from auction theory. A …rstprice sealed bid auction has bidders submit bids simultaneously to an auctioneer who awards the item to the highest bidder who then pays his bid. This is a very common auction form. A second-price sealed bid auction has bidders submit bids simultaneously to an auctioneer who awards the item to the highest bidder, just like before, but this time the winning bidder pays the second-highest price. To model the second-price auction, suppose that there are n bidders and 162 CHAPTER 12. INTEGRATION that bidder i values the item being auctioned at vi , which is independent of how much everyone else values the item. Bidders do not know their opponents’ valuations, but they do know the probability distribution of the opponents’ valuations. Bidder i must choose his bid bi . Let Fi (b) be the probability that the highest other bid faced by i, that is, the highest bid except for bi , is no larger than b. Then Fi (b) is a probability distribution function, and its density function is fi (b). Bidder i’s expected payo¤ is Z bi Vi (bi ) = (vi b)fi (b)db: 0 Let’s interpret this function. Bidder i wins if his is the highest bid, which occurs if the highest other bid is between 0 (the lowest possible bid) and his own bid bi . If the highest other bid is above bi bidder i loses and gets a payo¤ of zero. This is why the integral is taken over the interval [0; bi ]. If bidder i wins he pays the highest other bid b, which is distributed according to the density function fi (b). His surplus if he wins is vi b, his value minus how much he pays. Bidder i chooses the bid bi to maximize his expected payo¤ Vi (bi ). Since this is a maximization problem we should …nd the …rst-order condition: Z bi d 0 Vi (bi ) = (vi b)fi (b)db = 0: dbi 0 Notice that we are di¤erentiating with respect to bi , which shows up only as the upper endpoint of the integral. Using Leibniz’s rule we can evaluate this …rst-order condition: Z bi d 0 = (vi b)fi (b)db dbi 0 Z bi @ dbi d0 = [(vi b)fi (b)] db + (vi bi )fi (bi ) (vi 0)fi (0): dbi dbi 0 @bi The …rst term is zero because (vi b)fi (b) is not a function of bi , and so the partial derivative is zero. The second term reduces to (vi bi )fi (bi ) because dbi =dbi is simply one. The third term is zero because the derivative d0=dbi = 0. This leaves us with the …rst-order condition 0 = (vi bi )fi (bi ): 163 CHAPTER 12. INTEGRATION Since density functions take on only nonnegative values, the …rst-order condition holds when vi bi = 0, or bi = vi . In a second-price auction the bidder should bid his value. This result makes sense intuitively. Let bi be bidder i’s bid, and let b denote the highest other bid. Suppose …rst that bidder i bids more than his value, so that bi > vi . If the highest other bid is in between these, so that vi < b < bi , bidder i wins the auction but pays b vi more than his valuation. He could have avoided this by bidding his valuation, vi . Now suppose that bidder i bids less than his value, so that bi < vi . If the highest other bid is between these two, so that bi < b < vi , bidder i loses the auction and gets nothing. But if he had bid his value he would have won the auction and paid b < vi , and so he would have been better o¤. Thus, the best thing for him to do is bid his value. 12.4 Problems 1. Suppose that f (x) is the density function for a random variable distributed uniformly over the interval [2; 8]. (a) Compute Z 8 xf (x)dx 2 (b) Compute Z 8 x2 f (x)dx 2 2. Compute the following derivative: Z 2 d t tx2 dx dt t2 3. Find the following derivative: d dt Z 4t2 t2 x3 dx 3t 4. Let U (a; b) denote the uniform distribution over the interval [a; b]. Find conditions on a and b that guarantee that U (a; b) …rst-order stochastically dominates U (0; 1). CHAPTER 13 Moments 13.1 Mathematical expectation Let x~ be a random variable with density function f (x) and let u(x) be a real-valued function. The expected value of u(~ x) is denoted E[u(~ x)] and it is found by the following rules. If x~ is discrete taking on value xi with probability f (xi ) then X E[u(~ x)] = u(xi )f (xi ): i If x~ is continuous the expected value of u(~ x) is given by Z 1 u(x)f (x)dx: E[u(~ x)] = 1 Since integrals are for adding, as we learned in the last chapter, these formulas really do make sense and go together. 164 165 CHAPTER 13. MOMENTS The expectation operator E[ ] is linear, which means that E[au(~ x)] = aE[u(~ x)] E[u(~ x) + v(~ x)] = E[u(~ x)] + E[v(~ x)] 13.2 The mean The mean of a random variable x~ is = E[~ x], that is, it is the expected value of the function u(x) = x. Consider the discrete distribution with outcomes (4; 10; 12; 20) and corresponding probabilities (0:1; 0:2; 0:3; 0:4). The mean is E[~ x] = (4)(0:1) + (10)(0:2) + (12)(0:3) + (20)(0:4) = 14 13.2.1 Uniform distribution The mean of the uniform distribution over the interval [a; b] is (a + b)=2. If you don’t believe me, draw it. To …gure it out from the formula, compute Z b 1 x E[~ x] = dx b a a Z b 1 xdx = b a a b 1 2 x = b a 2 a b 2 a2 1 = b a 2 b+a = : 2 1 13.2.2 Normal distribution The mean of the general normal distribution is the parameter . Recall that the normal density function is 1 f (x) = p e 2 (x )2 =2 2 : 166 CHAPTER 13. MOMENTS The mean is Z x 2 2 p e (x ) =2 dx: 2 Use the change-of-variables formula y = x so that x = + y, (x )2 = 2 = y 2 , and dx = dy. Then we can rewrite Z x 2 2 p e (x ) =2 dx E[~ x] = Z 1 2 + y 2 p e y =2 dy = 2 Z11 Z 1 1 2 y 2 =2 p e dy + p = ye y =2 dy: 2 2 1 1 The …rst integral is the integral of the standard normal density, and like all densities its integral is 1. The second integral can be split into two parts: Z 1 Z 0 Z 1 2 y 2 =2 y 2 =2 ye y =2 dy: ye dy + ye dy = E[~ x] = 0 1 1 Use the change of variables y = z in the …rst integral on the right-hand side. Then y 2 = z 2 and dy = dz, so Z 1 Z 0 2 y 2 =2 ze z =2 dz ye dy = 0 1 Plugging this back into the expression above it yields Z 1 Z 1 Z 1 z 2 =2 y 2 =2 ye ze dz + ye dy = 0 1 y 2 =2 dy: 0 But both integrals on the right-hand side are the same, so the expression is zero. Thus, we get E[~ x] = . 13.3 Variance The variance of the random variable x~ is E[(~ x )2 ], where = E[~ x] is the 2 mean of the random variable. The variance is denoted . Note that E[(~ x )2 ] = = = = E[~ x2 2 x~ + 2 ] E[~ x2 ] 2 E[~ x] + 2 2 E[~ x] 2 + 2 2 E[~ x2 ] . 2 167 CHAPTER 13. MOMENTS We can …nd the variance of the discrete random variable used in the preceding section. The outcomes were (4; 10; 12; 20) and the corresponding probabilities were (0:1; 0:2; 0:3; 0:4). The mean was 14. The variance is )2 ] = (0:1)(4 14)2 + (0:2)(10 14)2 + (0:3)(12 14)2 + (0:4)(20 14)2 = (0:1)(100) + (0:2)(16) + (0:3)(4) + (0:4)(36) = 28:8 E[(~ x We can also …nd it using the alternative formula: E[~ x2 ] 2 = (0:1)(42 ) + (0:2)(102 ) + (0:3)(12)2 + (0:4)(202 ) 142 : You should be able to show that V ar(a~ x) = a2 V ar(~ x): p x )2 ], The standard deviation of the random variable x~ is E[(~ which means that the standard deviation is simply . It is the square root of the variance. 13.3.1 Uniform distribution The variance of the uniform distribution can be found from Z b 2 x 2 E[~ x] = dx a a b 1 b 1 3 x b a 3 a b 3 a3 1 = b a 3 = 168 CHAPTER 13. MOMENTS and note that b3 a3 = (b 2 13.3.2 a)(b2 + ab + a2 ). Consequently, 2 = E[~ x2 ] b2 + ab + a2 (b + a)2 = 3 4 2 2 4b + 4ab + 4a 3b2 6ab = 12 b2 2ab + a2 = 12 (b a)2 : = 12 3a2 Normal distribution The variance of the standard normal distribution is 1. Let’s take that on faith. The variance of the general normal distribution had better be the parameter 2 . To make sure, compute Z 1 (x )2 (x )2 =2 2 2 p e dx E[(~ x ) ]= 2 1 Using the same change-of-variables trick as before, we get Z 1 (x )2 (x )2 =2 2 2 p e dx E[(~ x )] = 2 1 Z 1 ( + y )2 y2 =2 p = e dy 2 1 Z 1 y2 2 p e y =2 dy = 2 1 Z 1 2 y 2 p e y =2 dy: = 2 2 1 The integral is the variance of the standard normal, which we already said was 1. 13.4 Application: Order statistics Suppose that you make n independent draws from the random variable x~ with distribution function F (x) and density f (x). The value of the highest CHAPTER 13. MOMENTS 169 of them is a random variable, the value of the second highest is a random variable, and so on, for n random variables. The n-th order statistic is the expected value of the n-th highest draw. So, the …rst order statistic is the expected value of the highest of the n draws, the second order statistic is the expected value of the second highest of the n draws, and so on. We use order statistics in a variety of settings, but the most straightforward one is auctions. Think about a …rst-price sealed bid auction in which n bidders submit their bids simultaneously and then the highest bidder wins and pays her bid. The seller’s expected revenue, then, is the expected value of the highest of the n bids, which is the …rst order statistic. Now think about the second-price sealed-bid auction. In this auction n bidders submit their bids simultaneously, the highest bid wins, and the winner pays the second-highest bid. The seller’s expected revenue in this auction is the expected value of the second-highest of the n bids, which is the second order statistic. As a side note, order statistics have also played a role in cosmology, the study of the cosmos, and in particular they were used by Edwin Hubble. Hubble was clearly an overachiever. In set the Illinois state high school record for high jump. He was a Rhodes scholar. He was the …rst astronomer to use the giant 200-inch Hale telescope at Mount Palomar. He was honored with a 41 cent postage stamp. He has the Hubble Space Telescope named after him. Importantly for this story, though, he established that the universe extends beyond our galaxy, the Milky Way. This was a problem because we know that stars that are farther away are dimmer, but not all stars have the same brightness. So, we can’t tell whether a particular star is dim because it’s far away or because it’s just not very bright (econometricians would call this an identi…cation problem). Astronomers before Hubble made the heroic (that is, unreasonable) assumption that all starts were the same brightness and worked from there. Hubble used the milder assumption that the brightest star in every nebula (or galaxy, but they didn’t know the di¤erence at the time) is equally bright. In other words, he assumed that the …rst order statistic is the same for every nebula. We want to …nd the order statistics, and, in particular, the …rst and second order statistics. To do this we have to …nd some distributions. Think about the …rst order statistic. It is the expected value of the highest of the n draws, and the highest of the n draws is a random variable with a distribution. But what is the distribution? We must construct it from the underlying distribution F . 170 CHAPTER 13. MOMENTS Let G(1) (x) denote the distribution for the highest of the n values drawn independently from F (x). We want to derive G(1) (x). Remember that G(1) (x) is the probability that the highest draw is less than or equal to x. For the highest draw to be less than or equal to x, it must be the case that every draw is less than or equal to x. When n = 1 the probability that the one draw is less than or equal to x is F (x). When n = 2 the probability that both draws are less than or equal to x is (F (x))2 . And so on. When there are n draws the probability that all of them are less than or equal to x is (F (x))n , and so G(1) (x) = F n (x): From this we can get the density function by di¤erentiating G(1) with respect to x: g (1) (x) = nF n 1 (x)f (x): Note the use of the chain rule. This makes it possible to compute the …rst order statistic, since we know the distribution and density functions for the highest of n draws. We just take the expected value in the usual way: Z (1) s = xnF n 1 (x)f (x)dx: Example 12 Uniform distribution over (0; 1). We have F (x) = x on [0; 1], and f (x) = 1 on [0; 1]. The …rst order statistic is Z (1) s = xnF n 1 (x)f (x)dx Z 1 xnxn 1 dx = 0 Z 1 xn dx = n 0 n+1 1 x n+1 n : = n+1 = n 0 This answer makes some sense. If n = 1 the …rst order statistic is just the mean, which is 1=2. If n = 2 and the distribution is uniform so that the 171 CHAPTER 13. MOMENTS draws are expected to be evenly spaced, then the highest draw should be about 2=3 and the lowest should be about 1=3. If n = 3 the highest draw should be about 3=4, and so on. We also care about the second order statistic. To …nd it we follow the same steps, beginning with identifying the distribution of the second-highest draw. To make this exercise precise, we are looking for the probability that the second-highest draw is no greater than some number, call it y. There are a total of n + 1 ways that we can get the second-highest draw to be below y, and they are listed below: Event Draw 1 is above y and the rest are below y Draw 2 is above y and the rest are below y .. . Probability (1 F (y))F n 1 (y) (1 F (y))F n 1 (y) .. . Draw n is above y and the rest are below y All the draws are below y (1 F (y))F n 1 (y) F n (y) Let’s …gure out these probabilities one at a time. Regarding the …rst line, the probability that draws 2 through n are below y is the probability of getting n 1 draws below y, which is F n 1 (y). The probability that draw 1 is above y is 1 F (y). Multiplying these together yields the probability of getting draw 1 above y and the rest below. The probability of getting draw 2 above y and the rest below is the same, and so on for the …rst n rows of the table. In the last row all of the draws are below y, in which case both the highest and the second highest draws are below y. The probability of all n draws being below y is just F n (y), the same as when we looked at the …rst order statistic. Summing the probabilities yields the distribution of the second-highest draw: G(2) (y) = n(1 F (y))F n 1 (y) + F n (y): Multiplying this out and simplifying yields G(2) (y) = nF n 1 (y) (n 1)F n (y): The density function is found by di¤erentiating G(2) with respect to y: g (2) (y) = n(n 1)F n 2 (y)f (y) n(n 1)F n 1 (y)f (y): 172 CHAPTER 13. MOMENTS It can be rearranged to get g (2) (y) = n(n 1)(1 F (y))F n 2 (y)f (y): The second order statistic is the expected value of the second-highest draw, which is Z (2) s = yg (2) (y)dy Z = yn(n 1)(1 F (y))F n 2 (y)f (y)dy: Example 13 Uniform distribution over [0; 1]. Z (2) s = yn(n 1)(1 F (y))F n 2 (y)f (y)dy Z 1 yn(n 1)(1 y)y n 2 dy = 0 Z 1 y n 1 y n dy = n(n 1) 0 1 yn n(n = n(n 1) n 0 n(n 1) = (n 1) n+1 n 1 : = n+1 1) y n+1 n+1 1 0 If there are four draws uniformly dispersed between 0 and 1, the highest draw is expected to be at 3=4, the second highest at 2=4, and the lowest at 1=4. If there are …ve draws, the highest is expected to be at 4=5 and the second highest is expected to be at 3=5, and so on. 173 CHAPTER 13. MOMENTS 13.5 Problems 1. Suppose that the random variable x~ takes on the following values with the corresponding probabilities: Value Probability 7 .10 4 .23 2 .40 -2 .15 -6 .10 -14 .02 (a) Compute the mean. (b) Compute the variance. 2. The following table shows the probabilities for two random variable, one with density function f (x), and one with density function g(x). x 10 15 20 30 100 f (x) 0.15 0.5 0.05 0.1 0.2 g(x) 0.20 0.30 0.1 0.1 0.3 (a) Compute the means of the two variables. (b) Compute the variances of the two variables. (c) Compute the standard deviations of the two variables. 3. Consider the triangular density given by f (x) = 2x on the interval [0; 1]. (a) Find its distribution function F . (b) Verify that it is a distribution function, that is, and speci…cally for this case, that F is increasing, F (0) = 0, and F (1) = 1. (c) Find the mean. 174 CHAPTER 13. MOMENTS (d) Find the variance. 4. Consider the triangular density given by f (x) = [0; 4]. 1 x 8 on the interval (a) Find its distribution function F . (b) Verify that it satis…es the properties of a distribution function, that is, F (0) = 0, F (4) = 1, and F increasing. (c) Find the mean. (d) Find the variance. 5. Show that if the variance of x~ is where a is a scalar. 6. Show that if the variance of x~ is of y~ is 9 2x . 2 x 2 then the variance of a~ x is a2 and if y~ = 3~ x 2 , 1, then the variance 7. Suppose that the random variable x~ takes the value 6 with probability 1 and takes the value y with probability 21 . Find the derivative d 2 =dy, 2 where 2 is the variance of x~. 8. Let G(1) and G(2) be the distribution functions for the highest and second highest draws, respectively. Show that G(1) …rst-order stochastically dominates G(2) . CHAPTER 14 Multivariate distributions Multivariate distributions arise when there are multiple random variables. For example, what we normally refer to as "the weather" is comprised of several random variables: temperature, humidity, rainfall, etc. A multivariate distribution function is de…ned over a vector of random variables. A bivariate distribution function is de…ned over two random variables. In this chapter I restrict attention to bivariate distributions. Everything can be extended to multivariate distributions by adding more random variables. 14.1 Bivariate distributions Let x~ and y~ be two random variables. The distribution function F (x; y) is given by F (x; y) = P (~ x x and y~ y): It is called the joint distribution function. The function F (x; 1) is the probability that x~ x and y~ 1. The latter is sure to hold, and so F (x; 1) is the univariate distribution function for the random variable x~. 175 CHAPTER 14. MULTIVARIATE DISTRIBUTIONS 176 Similarly, the function F (1; y) is the univariate distribution function for the random variable y~. The density function depends on whether the random variables are continuous or discrete. If they are both discrete then the density is given by f (x; y) = P (~ x = x and y~ = y). If they are both continuous the density is given by @2 f (x; y) = F (x; y): @x@y This means that the distribution function can be recovered from the density using the formula Z Z y x 1 1 F (x; y) = 14.2 f (s; t)dsdt: Marginal and conditional densities Consider the following example with two random variables: x~ = 1 x~ = 2 x~ = 3 y~ = 1 0.1 0.2 0.1 y~ = 2 0.3 0.1 0.2 The random variable x~ can take on three possible values, and the random variable y~ can take on two possible values. The probabilities in the table are the values of the joint density function f (x; y). Now add a total row and a total column to the table: x~ = 1 x~ = 2 x~ = 3 fy~(y) y~ = 1 0.1 0.2 0.1 0.4 y~ = 2 0.3 0.1 0.2 0.6 fx~ (x) 0.4 0.3 0.3 1 CHAPTER 14. MULTIVARIATE DISTRIBUTIONS 177 The sum of the …rst column is the total probability that y~ = 1, and the sum of the second column is the total probability that y~ = 2. These are the marginal densities. For a discrete bivariate random variable (~ x; y~) we de…ne the marginal density of x~ by fx~ (x) = n X f (x; yi ) i=1 where the possible values of y~ are y1 ; :::; yn . For the continuous bivariate random variable we de…ne the marginal density of x~ by Z 1 f (x; y)dy: fx~ (x) = 1 From this we can recover the marginal distribution function of x~ by integrating with respect to x: Z x Z 1 Z x f (t; y)dydt: fx~ (t)dt = Fx~ (x) = 1 1 1 We have already discussed conditional probabilities. We would like to have conditional densities. From the table above, it is apparent that the conditional density of x~ given the realization of y~ is f (xjy) = f (x; y)=f (y). To see that this is true, look for the probability that x~ = 3 given y~ = 2. The probability that y~ = 2 is fy~(2) = 0:6. The probability that x~ = 3 and y~ = 2 is f (3; 2) = 0:2. The conditional probability is f (~ x = 3j~ y = 2) = f (3; 2)=fy~(2) = 0:2=0:6 = 1=3. So, the rule is just what we would expect in the discrete case. What about the continuous case? The same formula works: f (xjy) = f (x; y) : fy~(y) So, it doesn’t matter in this case whether the random variables are discrete or continuous for us to …gure out what to do. Both of these formulas require conditioning on a single realization of y~. It is possible, though, to de…ne the conditional density much more generally. Let A be an event, and let P be the probability measure over events. Then we can write the conditional density f (x; A) f (xjA) = P (A) 178 CHAPTER 14. MULTIVARIATE DISTRIBUTIONS where f (x; A) denotes the probability that both x and A occur, written as a density function. For example, if A = fx : x~ x0 g, so that A is the event that the realization of x~ is no greater than x0 , we know that P (A) = F (x0 ). Therefore ( f (x) if x x0 F (x0 ) f (xj~ x x0 ) = 0 if x > x0 At this point we have too many functions ‡oating around. table to help with notation and terminology. Function Density Distribution Notation f (x; y) or fx~;~y (x; y) F (x; y) or Fx~;~y (x; y) Formula Discrete: P P y~ y Univariate dist. F (x) Marginal density fx~ (x) x ~ x Continuous: Ry Rx 1 Conditional density f (xjy) or fx~jy (xjy) Here is a 1 f (x; y) f (s; t)dsdt F (x; 1) P Discrete: yRf (x; y) 1 Continuous: f (x; y)dy 1 f (x; y)=f (y) The random variables x~ and y~ are independent if fx~;~y (x; y) = fx~ (x)fy~(y). In other words, the random variables are independent if the bivariate density is the product of the marginal densities. Independence implies that f (xjy) = fx~ (x) and f (yjx) = fy~(y), so that the conditional densities and the marginal densities coincide. 14.3 Expectations Suppose we have a bivariate random variable (~ x; y~). Let u(x; y) be a realvalued function, in which case u(~ x; y~) is a univariate random variable. Then the expected value of u(~ x; y~) is XX E[u(~ x; y~)] = u(x; y)f (x; y) y x 179 CHAPTER 14. MULTIVARIATE DISTRIBUTIONS in the discrete case and E[u(~ x; y~)] = Z 1 1 Z 1 u(x; y)f (x; y)dxdy 1 in the continuous case. It is still possible to compute the means of the random variables x~ and y~ separately. We can do this using the marginal densities. So, for example, in the table above the mean of y~ is (0:4)(1) + (0:6)(2) = 1:6: A particularly important case is where u(x; y) = (x x )(y y ), where ~ and y is the mean of y~. The resulting expectation is x is the mean of x called the covariance of x~ and y~, and it is denoted xy Note that xx = Cov(~ x; y~) = E[(~ x y x )(~ y )]: is just the variance of x~. Also, it is easy to show that xy = E[~ xy~] x y. The quantity xy = xy x y is called the correlation coe¢cient between x~ and y~. The following theorems apply to correlation coe¢cients. Theorem 20 If x~ and y~ are independent then xy = xy = 0. Proof. When the random variables are independent, f (x; y) = fx~ (x)fy~(y). Consequently we can write xy = E[(~ x y x )(~ y )] = E[~ x x] E[~ y y )]: But each of the expectations on the right-hand side are zero, and the result follows. It is important to remember that the converse is not true: sometimes two variables are not independent but still happen to have a zero covariance. An example is given in the table below. One can compute that xy = 0 but note that f (2; 6) = 0 while fx~ (2) fy~(6) = (0:2)(0:4) 6= 0. 180 CHAPTER 14. MULTIVARIATE DISTRIBUTIONS y~ = 6 0.2 0 0.2 x~ = 1 x~ = 2 x~ = 3 Theorem 21 y~ = 8 0 0.2 0 1: xy Proof. Consider the random variable x~ variances cannot be negative, we have 2 x ty 2 0 y~ = 10 0.2 0 0.2 2t~ xy~ + t2 y~2 ] = E[~ x = E[~ x2 ] = 2 x 2 x t2 2y + t~ y , where t is a scalar. Because 2 x ( 2t y2] + t2 E[~ 2t 2 y x y + t2 2 y) 2t E[~ xy~] x y xy : Since this is true for any scalar t, choose t= xy : 2 y + xy 2 y Substituting gives us 0 2 x 0 2 x 2 xy 2 2 x y xy 2 2 y 2 xy 2 y xy 2 xy 2 y 1 1: x y The theorem says that the correlation coe¢cient is bounded between 1 and 1. If xy = 1 it means that the two random variables are perfectly correlated, and once you know the value of one of them you know the value of the other. If xy = 1 the random variables are perfectly negatively CHAPTER 14. MULTIVARIATE DISTRIBUTIONS 181 correlated. This contains just as much information as perfect correlation. If you know that x~ has attained its highest possible value and x~ and y~ are perfectly negatively correlated, then y~ must have attained its lowest value. Finally, if xy = 0 the two variables are perfectly uncorrelated (and possibly independent). 14.4 Conditional expectations When there are two random variables, x~ and y~, one might want to …nd the expected value of x~ given that y~ has attained a particular value or set of values. This would be the conditional mean. We can use the above table for an example. What is the expected value of x~ given that y~ = 8? x~ can only take one value when y~ = 8, and that value is 2. So, the conditional mean of x~ given that y~ = 8 is 2. The conditional mean of x~ given that y~ = 10 is also 2, but for di¤erent reasons this time. To make this as general as possible, let u(x) be a function of x but not of y. I will only consider the continuous case here; the discrete case is similar. The conditional expectation of u(~ x) given that y~ = y is given by Z 1 Z 1 1 u(x)f (xjy)dx = E[u(~ x)j~ y = y] = u(x)f (x; y)dx: fy~(y) 1 1 Note that this expectation is a function of y but not a function of x. The reason is that x is integrated out on the right-hand side, but y is still there. 14.4.1 Using conditional expectations - calculating the bene…t of search Consider the following search process. A consumer, Max, wants to buy a particular digital camera. He goes to a store and looks at the price. At that point he has three choices: (i) but the camera at that store, (ii) go to another store to check its price, or (iii) go back to a previous store and buy the camera there. Stores draw their prices independently from the distribution CHAPTER 14. MULTIVARIATE DISTRIBUTIONS 182 F (p) given by p p p p = = = = 200 190 180 170 with with with with probability probability probability probability 0:2 0:3 0:4 0:1 We want to answer the following question: If the lowest price so far is q, what is the expected bene…t from checking one more store? Let’s begin by answering this in the most straightforward way possible. Suppose that q = 200, so that the lowest price found so far is the worst possible price. If Max searches one more time there is a 10% chance of …nding a price of $170 and saving $30, a 40% chance of …nding a price of $180 and saving $20, a 30% chance of …ncing a price of $190 and saving only $10, and a 20% chance of …nding another store that charges the highest possible price of $200, in which case the savings are zero. The expected saving is (:1)(30) + (:4)(20) + (:3)(10) + (:2)(0) = 14. When q = 200, the expected bene…t of search is $14. Now suppose that q = 190, so that the best price found so far is $190. Max has a 10% chance of …nding a price of $170 and saving $20, a 40% chance of …nding a price of $180 and saving $10, a 30% chance of …nding the same price and saving nothing, and a 20% chance of …nding a higher price of $200, in which case he also saves nothing. The expected saving is (:1)(20) + (:4)(10) + (:3)(0) + (:2)(0) = 6. When the best price found so far is q = 190, the expected bene…t of search is $6. Finally, suppose that q = 180. Now there is only one way to improve, which comes by …nding a store that charges a price of $170, leading to a $10 saving. The probabiliyt of …nding such a store is 10%, and the expected saving from search is $1. So now we know the answers, and let’s use these answers to …gure out a general formula, speci…cally one involving conditional expectations. Note that when Max …nds a price of p and the best price so far is q, his bene…t is q p if the new price p is lower than the old price q. Otherwise the bene…t is zero because he would be better o¤ buying the item at a store he’s already found. This "if" statement lends itself to a conditional expectation. In particular, the "if" statement pertains to the conditional expectation E[q p~j~ p < q], where the expectation is taken over the random variable p. This epxression tells us what the average bene…t is provided that the bene…t is 183 CHAPTER 14. MULTIVARIATE DISTRIBUTIONS nonnegative. The actual expected bene…t is Prf~ p < qgE[q p~j~ p < q]; which is the probability that the bene…t is positive times the expected bene…t conditional on the bene…t being positive. Let’s make sure this works using the above example. In particular, let’s look at q = 190. The conditional expectation is E[190 p~j~ p < 190] = (:4)(190 = 6; 180) + (:1)(190 170) which is exactly what we found before. The conditional expectation lets us work with more complicated distributions. Suppose that prices are drawn independently from the uniform distribution over the interval [150; 200]. Let the corresponding distribution function be F (p) and the density function be f (p). The expected bene…t from searching at another store when the lowest price so far is q is Z q f (p) [q p] Prf~ p < qgE[q p~j~ p < q] = F (q) dp F (q) 150 Z q [q p]f (p)dp: = 150 To see why this works, look at the top line. The probability that p~ < q is simply F (q), because that is the de…nition of the distribution function. That gives us the …rst term on the right-hand side. For the second term, note that we are taking the expectation of q p, so that term is in brackets. To …nd the conditional expectation, we multiply by the conditional density which is the density of the random variable p divided by the probability that the conditioning event (~ p < q) occurs. We take the integral over the interval [150; q] because outside of this interval the value of the bene…t is zero. When we multiply the two terms on the right-hand side of the top line together, we …nd that the F (q) term cancels out, leaving us with the very simple bottom line. Using it we can …nd the net bene…t of searching at one more store when the best price so far is $182.99: Z 182:99 Z q 1 [182:99 p] dp = 10:883: [q p]f (p)dp = 50 150 150 CHAPTER 14. MULTIVARIATE DISTRIBUTIONS 14.4.2 184 The Law of Iterated Expectations There is an important result concerning conditional expectations. It is called the Law of Iterated Expectations, and it goes like this. Ey [Ex [u(~ x)j~ y = y]] = Ex [u(~ x)]: It’s a complicated statement, so let’s look at what it means. The inside of the left-hand side is the conditional expectation of u(~ x) given that y~ takes some value y. As we have already learned, this is a function of y but not a function of x. Let’s call it v(y), and v(~ y ) is a random variable. So now let’s take the expectation of v(~ y ). The Law of Iterated Expectations says that E[v(~ y )] = Ex [u(~ x)]. Another way of looking at it is taking the expectation of a conditional expectation. Doing that removes the conditional part. The best thing to do here is to look at an example to see what’s going on. Let’s use one of our previous examples: x~ = 1 x~ = 2 x~ = 3 fy~(y) y~ = 1 0.1 0.2 0.1 0.4 y~ = 2 0.3 0.1 0.2 0.6 fx~ (x) 0.4 0.3 0.3 1 Begin by …nding the conditional expectations E[~ xj~ y = 1] and E[~ xj~ y = 2]. We get E[~ xj~ y = 1] = 2 and E[~ xj~ y = 2] = 11=6. Now take the expectation over y to get Ey [Ex [u(~ x)j~ y = y]] = fy~(1) E[~ xj~ y = 1] + fy~(2) E[~ xj~ y = 2] = (0:4)(2) + (0:6)(11=6) = 1:9: Now …nd the unconditional expectation of x~. It is Ex [~ x] = fx~ (1) 1 + fx~ (2) 2 + fx~ (3) 3 = (0:4)(1) + (0:3)(2) + (0:3)(3) = 1:9: It works. CHAPTER 14. MULTIVARIATE DISTRIBUTIONS 185 Now let’s look at it generally using the continuous case. Begin with Z 1 u(x)fx~ (x)dx Ex [u(~ x)] = 1 Z 1 Z 1 u(x) = f (x; y)dy dx 1 1 Z 1Z 1 u(x)f (x; y)dydx: = 1 1 Note that f (x; y) = f (xjy)fy~(y), so we can rewrite the above expression Z 1Z 1 u(x)f (xjy)fy~(y)dydx Ex [u(~ x)] = 1 1 Z 1 Z 1 = u(x)f (xjy)dx fy~(y)dy 1 1 Z 1 Ex [u(~ x)j~ y = y]fy~(y)dy = 1 = Ey [Ex [u(~ x)j~ y = y]]: 14.5 Problems 1. There are two random variables, x~ and y~, with joint density f (x; y) given by the following table. f (x; y) x~ = 1 x~ = 2 x~ = 3 x~ = 4 y~ = 10 :04 :07 :02 :01 y~ = 20 0 0 :11 :12 y~ = 30 :20 :18 :07 :18 (a) Construct a table showing the distribution function F (x; y). (b) Find the univariate distributions Fx~ (x) and Fy~(y). (c) Find the marginal densities fx~ (x) and fy~(y). (d) Find the conditional density f (xj~ y = 20). (e) Find the mean of y~. (f) Find the mean of x~ conditional on y~ = 20. CHAPTER 14. MULTIVARIATE DISTRIBUTIONS 186 (g) Are x~ and y~ independent? (h) Verify that the Law of Iterated Expectations works. 2. There are two random variables, x~ and y~, with joint density given by the following table: f (x; y) x~ = 1 x~ = 2 x~ = 3 x~ = 4 y~ = 3 0.03 0.02 0.05 0.07 y~ = 8 0.02 0.12 0.01 0.11 y~ = 10 0.20 0.05 0.21 0.11 (a) Construct a table showing the distribution function F (x; y). (b) Find the univariate distributions Fx~ (x) and Fy~(y). (c) Find the marginal densities fx~ (x) and fy~(y). (d) Find the conditional density f (yj~ x = 3). (e) Find the means of x~ and y~. (f) Find the mean of x~ conditional on y~ = 3. (g) Are x~ and y~ independent? (h) Find V ar(~ x) and V ar(~ y ). (i) Find Cov(~ x; y~). (j) Find the correlation coe¢cient between x~ and y~. (k) Verify the Law of Iterated Expectations for …nding Ex [~ x] =.Ey [Ex [~ xjy]]: 3. Let F (x) be the uniform distribution over the interval [a; b], and suppose that c 2 (a; b). Show that F (xjx c) is the uniform distribution over [a; c]. 4. Consider the table of probabilities below: f (x; y) x~ = 1 x~ = +1 y~ = 10 0.1 0.3 y~ = 20 a b What values must a and b take for x~ and y~ to be independent? CHAPTER 15 Statistics 15.1 Some de…nitions The set of all of the elements about which some information is desired is called the population. Examples might be the height of all people in Knoxville, or the ACT scores of all students in the state, or the opinions about Congress of all people in the US. Di¤erent members of the population have di¤erent values for the variable, so we can treat the population variable as a random variable x~. So far everything we have done in probability theory is about the population random variable. In particular, its mean is and its variance is 2 . A random sample from a population random variable x~ is a set of independent, identically distributed (IID) random variables x~1 ; x~2 ; :::; x~n , each of which has the same distribution as the parent random variable x~. The reason for random sampling is that sometimes it is too costly to measure all of the elements of a population. Instead, we want to infer properties of the entire population from the random sample. This is statistics. Let x1 ; :::; xn be the outcomes of the random sample. A statistic is a 187 188 CHAPTER 15. STATISTICS function of the outcomes of the random sample which does not contain any unknown parameters. Examples include the sample mean and the sample variance. 15.2 Sample mean The sample mean is given by x= x1 + ::: + xn : n Note that we use di¤erent notation for the sample mean (x) and the population mean ( ). The expected value of the sample mean can be found as follows: n 1X E[x] = E[~ xi ] n i=1 n 1X = n i=1 1 (n ) = : n = So, the expected value of the sample mean is the population mean. The variance of the sample mean can also be found. To do this, though, let’s …gure out the variance of the sum of two independent random variables x~ and y~. Theorem 22 Suppose that x~ and y~ are independent. Then V ar(~ x + y~) = V ar(~ x) + V ar(~ y ). Proof. Note that (x x +y y) 2 = (x x) 2 + (y y) 2 + 2(x x )(y y ): Take the expectations of both sides to get 2 V ar(~ x + y~) = E[(~ x ~ x+y y) ] 2 2 = E[(~ x y x x ) ] + E[(~ y ) ] + 2E[(~ = V ar(~ x) + V ar(~ y ) + 2Cov(~ x; y~): y x )(~ y )] 189 CHAPTER 15. STATISTICS But, as shown in Theorem 20, since x~ and y~ are independent, Cov(~ x; y~) = 0, and the result follows. We can use this theorem to …nd the variance of the sample mean. Since a random sample is a set of IID random variables, the theorem applies. Also, recall that V ar(a~ xi ) = a2 V ar(~ xi ). So, ! n 1X V ar(x) = V ar x~i n i=1 n 1 X V ar(~ xi ) = n2 i=1 = n 2 n2 2 = n : This is a really useful result. It says that the variance of the sample mean around the population mean shrinks as the sample size becomes larger. So, bigger samples imply better …ts, which we all knew already but we didn’t know why. 15.3 Sample variance We are going to use two di¤erent pieces of notation here. One is n 1X m = (xi n i=1 2 and the other is 1 2 s = n 1 n X (xi x)2 x)2 i=1 Both of these can be interpreted as the estimates of variance, and many scienti…c calculators compute both of them. What you should remember is to use m2 as the computed variance when the random sample coincides with the entire population, and to use s2 when the random sample is a subset of the population. In other words, you will almost always use s2 , and we refer 190 CHAPTER 15. STATISTICS to s2 as the sample variance. But, m2 is useful for what we are about to do. We want to …nd the expected value of the sample variance s2 . It is easier to …nd the expected value of m2 and note that s2 = n n 1 m2 : We get # " n X 1 E[m2 ] = E (~ xi x)2 n " i=1 # n X 1 E E[x2 ] x~i 2 = n i=1 which follows from a previous manipulation of variance: V ar(~ x) = E[(~ x 2 2 2 ) ] = E[~ x] . Rearranging that formula and applying it to the sample mean tells us that E[x2 ] = E[x]2 + V ar(x); so n 1X E[~ x2i ] n i=1 E[m2 ] = ! E[x]2 V ar(x): But we already know some of these values. We know that E[x] = V ar(x) = 2 =n. Finally, note that since x~i has mean and variance have E[~ x2i ] = 2 + 2 : Plugging this all in yields n 1X ( n i=1 E[m2 ] = = ( = 2 n + 1 n 2 2 ) 2 2 + 2 ) ! 2 2 n 2 n 2 and , we 191 CHAPTER 15. STATISTICS Now we can get the mean of the sample variance s2 : n E[s2 ] = E = = = n n n 1 n n 2 1 m2 E[m2 ] n 1 1 2 n : The reason for using s2 as the sample variance instead of m2 is that s2 has the right expected value, that is, the expected value of the sample variance is equal to the population variance. We have found that E[x] = and E[s2 ] = 2 . Both x and s2 are statistics, because they depend only on the observed values of the random sample and they have no unknown parameters. They are also unbiased because their expected values are equal to the population parameters. Unbiasedness is an important and valuable property. Since we use random samples to learn about the characteristics of the entire population, we want statistics that match, in expectation, the parameters of the population distribution. We want the sample mean to match the population mean in expectation, and the sample variance to match the population variance in expectation. Is there any intuition behind dividing by n 1 in the sample variance instead of dividing by m? Here is how I think about it. The random sample has n observations in it. We only need one observation to compute a sample mean. It may not be a very good or precise estimate, but it is still an estimate. Since we can use the …rst observation to compute a sample mean, we can use all of the data to compute all of the data to compute a sample mean. This may seem cryptic and obvious, but now think about what we need in order to compute a sample variance. Before we can compute the sample variance, we need to compute the sample mean, and we need at least one observation to do this. That leaves us with n 1 observations to compute the sample variance. The terminology used in statistics is degrees of freedom. With n observations we have n degrees of freedom when we compute the sample mean, but we only have n 1 degrees of freedom when we compute the sample variance because one degree of freedom was used to compute the sample mean. In both calculations (sample mean and sample variance) we divide by the number of degrees of freedom, n for the sample mean x and n 1 for the sample variance s2 . 192 CHAPTER 15. STATISTICS 15.4 Convergence of random variables In this section we look at what happens when the sample size becomes in…nitely large. The results are often referred to as asymptotic properties. We have two main results, both concerning the sample mean. One is called the Law of Large Numbers, and it says that as the sample size grows without bound, the sample mean converges to the population mean. The second is the Central Limit Theorem, and it says that the distribution of the sample mean converges to a normal distribution, regardless of whether the population is normally distributed or not. 15.4.1 Law of Large Numbers Let xn be the sample mean from a sample of size n. The basic law of large numbers is xn ! when n ! 1: The only remaining issue is what that convergence arrow means. The Weak Law of Large Numbers states that for any " > 0 lim P (jxn n!1 j < ") = 1. To understand this, take any small positive number ". What is the probability that the sample mean xn is within " of the population mean? As the sample size grows, the sample mean should get closer and closer to the population mean. And, if the sample mean truly converges to the population mean, the probability that the sample mean is within " of the population mean should get closer and closer to 1. The Weak Law says that this is true no matter how small " is. This type of convergence is called convergence in probability, and it is written P xn ! when n ! 1: The Strong Law of Large Numbers states that P lim xn = n!1 = 1: This one is a bit harder to understand. It says that the sample mean is almost sure to converge to the population mean. In fact, this type of convergence is called almost sure convergence and it is written a:e: xn ! when n ! 1: 193 CHAPTER 15. STATISTICS 15.4.2 Central Limit Theorem This is the most important theorem in asymptotic theory, and it is the reason why the normal distribution is so important to statistics. Let N ( ; 2 ) denote a normal distribution with mean and variance 2 . Let (z) be the distribution function (or cdf) of the normal distribution with mean 0 and variance 1 (or the standard normal N (0; 1)). To state it, compute the standardized mean: xn Zn = p E[xn ] V ar(xn ) : We know some of these values: E[xn ] = and V ar(xn ) = get xn p : Zn = = n 2 =n. Thus we The Central Limit Theorem states that if V ar(~ xi ) = if the population random variable has …nite variance, then 2 < 1, that is, lim P (Zn n!1 z) = (z): In words, the distribution of the standardized sample mean converges to the standard normal distribution. This kind of convergence is called convergence in distribution. 15.5 Problems 1. You collect the following sample of size n = 12: 10; 4; 1; 3; 2; 8; 6; 8; 6; 1; 5; 10 Find the sample mean and sample variance. CHAPTER 16 Sampling distributions Remember that statistics, like the mean and the variance of a random variable, are themselves random variables. So, they have probability distributions. We know from the Central Limit Theorem that the distribution of the sample mean converges to the normal distribution as the sample size grows without bound. The purpose of this chapter is to …nd the distributions for the mean and other statistics when the sample size is …nite. 16.1 Chi-square distribution The chi-square distribution turns out to be fundamental for doing statistics because it is closely related to the normal distribution. Chi-square random variables can only have nonnegative values. It turns out to be the distribution you get when you square a normally distributed random variable. The density function for the chi-square distribution with n degrees of freedom is n x (x=2) 2 1 e 2 f (x) = 2 (n=2) 194 195 CHAPTER 16. SAMPLING DISTRIBUTIONS y 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0 1 2 3 4 5 x Figure 16.1: Density for the chi-square distribution. The thick line has 1 degree of freedom, the thin line has 3, and the dashed line has 5. R1 where (a) is the Gamma function de…ned as (a) = 0 y a 1 e y dy for a > 0. 2 We use the notation y~ n to denote a chi-square random variable with n degrees of freedom. The density for the chi-square distribution with di¤erent degrees of freedom is shown in Figure 16.1. The thick line is the density with 1 degree of freedom, the thin line has 3 degrees of freedom, and the dashed line has 5. Changing the degrees of freedom radically changes the shape of the density function. One strange thing about the chi-square distribution is its mean and variance. The mean of a 2n random variable is n, the number of degrees of freedom, and the variance is 2n. The relationship between the standard normal and the chi-square distribution is given by the following theorem. Theorem 23 If x~ has the standard normal distribution, then the random variable x~2 has the chi-square distribution with 1 degree of freedom. 196 CHAPTER 16. SAMPLING DISTRIBUTIONS Proof. The distribution function for the variable y~ = x~2 is Fy~(y) = = = = P (~ y y) P (~ x2 y) p p P( y x y) p y) 2P (0 x where the last equality follows from the fact that the standard normal distribution is symmetric around its mean of zero. From here we can compute Z py Fy~(y) = 2 fx~ (x)dx 0 where fx~ (x) is the standard normal density. Using Leibniz’s’ rule, di¤erentiate this with respect to y to get the density of y~: fy~(y) = 2fx~ ( p y) where the last term is the derivative of formula for the normal density to get 1 fy~(y) = p e 2 y=2 y p 1=2 1 p 2 y y with respect to y. Plug in the (y=2) 1=2 e p = 2 y=2 : This looks exactly like the p formula for the density of 21 except for the de, and the formula is complete. nominator. But, (1=2) = Both the normal distribution and the chi-square distribution have the property that they are additive. That is, if x~ and y~ are independent normally distributed random variables, then z~ = x~ + y~ is also normally distributed. If x~ and y~ are independent chi-square random variables with nx and ny degrees of freedom, respectively, then z~ = x~ + y~ is has a chi-square distribution with nx + ny degrees of freedom. 16.2 Sampling from the normal distribution We use the notation x~ N ( ; 2 ) to denote a random variable that is distributed normally with mean and variance 2 . Similarly, we use the notation CHAPTER 16. SAMPLING DISTRIBUTIONS 197 2 x~ ~ has the chi-square distribution with n degrees of freedom. n when x The next theorem describes the distribution of the sample statistics of the standard normal distribution. Theorem 24 Let x1 ; :::; xn be an IID random sample from the standard normal distribution N (0; 1). Then (a) x N (0; n1 ): P (b) (xi x)2 2 n 1: (c) The random variables x and P (xi x)2 are statistically independent. The above theorem relates to random sampling from a standard normal distribution. If x1 ; :::; xn are an IID random sample from the general normal distribution N ( ; 2 ), then 2 x Also, P (xi N( ; x)2 2 n ): 2 n 1: To …gure out the distribution of the sample variance, …rst note that since the sample variance is P (xi x)2 2 s = n 1 2 2 and E[s ] = , we get that (n 1)s2 2 2 n 1: This looks more useful than it really is. In order to get it you need to know the population variance, 2 , in which case you don’t really need the sample variance, s2 . The next section discusses the sample distribution of s2 when 2 is unknown. The properties to take away from this section are that the sample mean of a normal distribution has a normal distribution, and the sample variance of a normal distribution has a chi-square distribution (after multiplying by (n 1)= 2 ). 198 CHAPTER 16. SAMPLING DISTRIBUTIONS 16.3 t and F distributions In econometrics the two most frequently encountered distributions are the t distribution and the F distribution. To brie‡y say where they arise, in econometrics one often runs a linear regression of the form yt = b0 + b1 x1;t + ::: + bk xk;t + ~"t where t is the observation number, xi;t is an observation of the i-th explanatory variable, and ~"t N (0; 2 ). One then estimates the coe¢cients b0 ; :::; bk in ways we have described earlier in this book. One uses a t distribution to test whether the individual coe¢cients b0 ; :::; bk are equal to zero. If the test rejects the hypothesis, that explanatory variable has a statistically signi…cant impact on the dependent variable. The F test is used to test if linear combinations of the coe¢cients are equal to zero. Let’s start with the t distribution. Graphically, its density looks like the normal density but with fatter tails, as in Figure 16.2. To get a t distribution, let x~ have a standard normal distribution and let y~ have a chisquare distribution with n degrees of freedom. Assume that x~ and y~ are independent. Construct the random variable t~ according to the formula x t= : (y=n)1=2 Then t~ has a t distribution with n degrees of freedom. In shorthand, t~n N (0; 1) p : 2 =n n Now look back at Theorem 24. If we sample from a standard normal, the sample mean has an N (0; 1) distribution and the sum of the squared deviations has a 2n distribution. So, we get a t distribution when we use the sample mean in the numerator and something like the sample variance in the denominator. The trick is to …gure out exactly what we need. If x~ N ( ; 2 ), then we know that x N ( ; 2 =n), in which case x p N (0; 1): = n P 2 We also know that ( (xi x)2 ) = 2 n 1 . Putting this all together yields t~ = r xp = n P (xi x)2 2 =(n 1) x =p s2 =n : (16.1) 199 CHAPTER 16. SAMPLING DISTRIBUTIONS y 0.4 0.3 0.2 0.1 -5 -4 -3 -2 -1 0 1 2 3 4 5 x Figure 16.2: Density for the t distribution. The thick curve has 1 degree of freedom and the thin curve has 5. The dashed curve is the standard normal density. The random variable t~ has a t distribution with n 1 degrees of freedom. Also notice that it does not depend on 2 . It does depend on , but we will discuss the meaning of this more in the next chapter. The statistic computed in expression (16.1) is commonly referred to as a t-statistic. Like the t distribution, the F distribution is a ratio of two other distributions. In this case it is the ratio of two chi-square distributions. The formula is 2 m =m ; Fm;n 2 =n n and the density function is shown in Figure 16.3. Because chi-square distributions assign positive probability only to non-negative outcomes, F distributions also assign positive probability only to non-negative outcomes. The F distribution and the t distribution are related. If t~n has the t distribution with n degrees of freedom, then (t~n )2 F1;n : Applying this to expression (16.1) tells us that the following sample statistic 200 CHAPTER 16. SAMPLING DISTRIBUTIONS y 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 1 2 3 4 5 x Figure 16.3: Density function for the F distribution. The thick curve is F2;20 , the dashed curve is F5;20 , and the thin curve is F10;20 . has an F distribution: (x )2 F~ = s2 =n 16.4 F1;n 1 : Sampling from the binomial distribution Recall that the binomial distribution is used for computing the probability of getting no more than s successes from n independent trials when the probability of success in any trial is p. A series of random trials is going to result in a sequence of successes or failures, and we can use the random variable x~ to capture this. Let xi = 1 if there Pwas a success in trial i and xi = 0 if there was a failure P in trial i. Then ni=1 xi is the number of successes in n trials, and x = ( xi ) =n is the average number of successes. Notice that x is also the sample frequency, that is, the fraction of successes in the sample. The sample frequency has the following properties: E[x] = p V ar(x) = p(1 p) n : CHAPTER 17 Hypothesis testing The tool of statistical analysis has two primary uses. One is to describe data, and we do this using such things as the sample mean and the sample variance. The other is to test hypotheses. Suppose that you have, for example, a sample of UT graduates and a sample of Vanderbilt graduates, both from the class of 2002. You may want to know whether or not the average UT grad makes more than the national average income, which was about $35,700 in 2006. You also want to know if the two classes have the same income. You would perform hypothesis tests to either support or reject the hypotheses that UT grads have higher average earnings than the national average and that both UT grads and Vanderbilt grads have the same average income. In general, hypothesis testing involves the value of some parameter that is determined by the data. There are two types of tests. One is to determine if the realized value of is in some set 0 . The other is to compute two di¤erent values of from two di¤erent samples and determine if they are the same or if one is larger than the other. 201 202 CHAPTER 17. HYPOTHESIS TESTING 17.1 Structure of hypothesis tests The …rst part of testing a hypothesis is forming one. In general a hypothesis takes the form of 2 0 , where 0 is a nonempty set of values for . It could be a single value, or it could be a range of values. The statement 2 0 is called the null hypothesis. The alternative hypothesis is that 2 = 0. We typically write these as H0 (Null hypothesis): 2 0 vs. H1 (Alternative hypothesis): 2 = 0. The form of the alternative hypothesis is determined completely by the form of the null hypothesis. So, if H0 is = 0 , H1 is 6= 0 . If H0 is > 0 . And so on. 0 , H1 is Another issue in hypothesis testing is that hypotheses can be rejected but they cannot be accepted. So, you can establish that something is false, but not that something is true. Because of this, empirical economists often make the null hypothesis something they would like to be false. If they can reject the null hypothesis, that lends support to the alternative hypothesis. For example, if one thinks that the variance of returns to the Dow-Jones Industrial Average is smaller than the variance of returns to the S&P 500 index, one would form the null hypothesis that the variance is at least as great for the Dow and then try to reject it. When running linear regressions, one tests the null hypothesis that the coe¢cients are zero. One uses a statistical test to either reject or support the null hypothesis. The nature of the test is as follows. First we compute a test statistic for . Let’s call it T . For example, if is the mean of the population distribution, T would be the sample mean. As we know, the value of T is governed by a random process. The statistical test identi…es a range of values A for the random variable T such that if T 2 A the null hypothesis is "accepted" and if T 2 = A the null hypothesis is rejected. The set A is called the critical region, and it is important to note that A and 0 are two completely di¤erent things. For example, a common null hypothesis is H0 : = 0. In that case 0 = f0g. But, we do not reject the null hypothesis if T is anything but zero, because then we would reject the hypothesis with probability 1. Instead, we reject the hypothesis if T is su¢ciently far from zero, or, in our new terminology, if T is outside of the critical region A. CHAPTER 17. HYPOTHESIS TESTING 203 Statistical tests can have errors because of the inherent randomness. It might be the case that 2 0 , so that the null hypothesis is really true, but T 2 = A so we reject the null hypothesis. Or, it might be the case that 2 = 0 so that the null hypothesis is really false, but T 2 A and we "accept" it anyway. The possible outcomes of the test are given in the table below. True value of parameter 2 0 2 = 0 Value of test statistic T T 2A T 2 =A Correctly "accept" null Incorrectly reject null Type I error Incorrectly "accept" null Correctly reject null Type II error A type I error occurs when one rejects a true null hypothesis. A type II error occurs when a false null hypothesis is not rejected. A problem arises because reducing the probability of a type I error generally increases the probability of a type II error. After all, reducing the probability of a type I error means rejecting the hypothesis less often, whether it is true or not. Let F (zj ) be the distribution of the test statistic z conditional on the value of the parameter . The entire previous chapter was about these distributions. If the null hypothesis is really true, the probability that the null is "accepted" is F (Aj 2 0 ). This is called the con…dence level. This probability of a type I error is 1 F (Aj 2 0 ). This probability is called the signi…cance level. The standard is to use a 5% signi…cance level, but 10% and 1% signi…cance levels are also reported. The 5% signi…cance level corresponds to a 95% con…dence level. I usually interpret the con…dence level as the level of certainty with which the null hypothesis is false when it is rejected. So, if I reject the null hypothesis with a 95% con…dence level, I am 95% sure that the null hypothesis is really false. Here, then, is the "method of proof" entailed in statistical analysis. Think of a statement you want to be true. Make this the alternative hypothesis. The null hypothesis is therefore a statement you would like to reject. Construct a test statistic related to the null hypothesis. Reject the null hypothesis if you are 95% sure that it is false given the value of the test statistic. CHAPTER 17. HYPOTHESIS TESTING 204 For example, suppose that you want to test the null hypothesis that the mean of a normal distribution is 0. This makes the hypothesis H0 : H1 =0 vs. : 6= 0 You take a sample x1 ; :::; xn and compute the sample mean x and sample variance s2 . Construct the test statistic x (17.1) t= p s2 =n which we know from Chapter 16 has a t distribution with n 1 degrees of freedom. In this case is the hypothesized mean, which is equal to zero. Now we must construct a critical range A for the test statistic t. Our critical range will be an interval around zero so that we reject the null hypothesis if t is too far from zero. We call this interval the 95% con…dence interval, and it takes the form (tL ; tH ). Let Tn 1 (t) be the distribution function for the t distribution with n 1 degrees of freedom. Then the endpoints of the con…dence interval satisfy Tn 1 (tL ) = 0:025 Tn 1 (tH ) = 0:975 The …rst line says that the probability that t is lower than tL is 2.5%, and the second says that the probability that t is higher than tH is 2.5%. Combining these means that the probability that the test statistic t is outside of the interval (tL ; tH ) is 5%. All of this is shown in Figure 17.1. The probability that t tL is 2.5%, as shown by the shaded area. The probability that t tH is also 2.5%. The probability that t is between these two points is 95%, and so the interval (tL ; tH ) is the 95% con…dence interval. The hypothesis is rejected if the value of t lies outside of the con…dence interval. Another way to perform the same test is to compute Tn 1 (t), where t is the test statistic given in (17.1) above. Reject the hypothesis if Tn 1 (t) < 0:025 or Tn 1 (t) > 0:975: If the …rst of these inequalities hold, the value of the test statistic is outside of the con…dence interval and to the left, and if the one on the right holds 205 CHAPTER 17. HYPOTHESIS TESTING y 2.5% area 2.5% area tL tH 0 x 95% confidence interval Figure 17.1: Two-tailed hypothesis testing the test statistic is outside of the con…dence interval and to the right. The null hypothesis cannot be rejected if 0:025 Tn 1 (t) 0:975. 17.2 One-tailed and two-tailed tests The tests described above are two-tailed tests, because rejection is based on the test statistic lying in one of the two tails of the distribution. Twotailed tests are used when the null hypothesis is an equality hypothesis, so that it is violated if the test statistic is either too high or too low. A one-tailed test is used for inequality-based hypotheses, such as the one below: H0 : H1 0 vs. : <0 In this case the null hypothesis is rejected if the test statistic is both far from zero and negative. Large test statistics are compatible with the null hypothesis as long as they are positive. This contrasts with the two-sided tests where large test statistics led to rejection regardless of the sign. 206 CHAPTER 17. HYPOTHESIS TESTING To test the null hypothesis that the mean of a normal distribution is nonnegative, compute the test statistic given in (17.1). We know from Chapter 16 that it has the t distribution with n 1 degrees of freedom. Letting Tn 1 (t) be the cdf of the t distribution with n 1 degrees of freedom, we reject the null hypothesis at the 95% con…dence level (or 5% signi…cance level) if Tn 1 (t) < 0:05: There are two di¤erences between the one-tailed criterion for rejection and the two-tailed criterion in the previous section. One di¤erence is that the one-tailed criterion can only be satis…ed one way, with Tn 1 (t) small, while the two-tailed criterion can be satis…ed two ways, with Tn 1 (t) either close to zero or close to one. The second di¤erence is that the one-tailed criterion has a cuto¤ point of 0.05, while the two-tailed criterion has a lower cuto¤ point half as big at 0.025. The reason for this is that the two-tailed test splits the 5% probability mass equally between the two tails, while the one-tailed criterion puts the whole 5% in the lower tail. The following table gives the rules for the one-tailed and two-tailed tests with signi…cance level and con…dence level 1 . The test statistic is z with distribution function G(z), and the hypotheses concern some parameter . Type of test Hypothesis H0 : = 0 Two-tailed vs. H1 : 6= 0 H0 : 0 Upper one-tailed vs. H1 : > 0 H0 : 0 Lower one-tailed vs. H1 : < 0 Reject H0 if G(z) < 2 or G(z) > 1 2 G(z) > 1 G(z) < p-value 2[1 1 G(jzj)] G(z) G(z) The p-value can be thought of as the exact signi…cance level for the test. The null hypothesis is rejected if the p-value is smaller than , the desired signi…cance level. 207 CHAPTER 17. HYPOTHESIS TESTING 17.3 Examples 17.3.1 Example 1 The following sequence is a random sample from the distribution N ( ; 2 ). The task is to test hypotheses about . The sequence is: 56, 74, 55, 66, 51, 61, 55, 48, 48, 47, 56, 57, 54, 75, 49, 51, 79, 59, 68, 72, 64, 56, 64, 62, 42. Test the null hypothesis that = 65. This is a two-tailed test based on the statistic in equation (17.1). We compute x = 58:73, s = 9:59, and n = 25. We get x t= p s2 =n = 58:73 65 = 9:59=5 3:27: The next task is to …nd the p-value using T24 (t), the t distribution with n 1 = 24 degrees of freedom. Excel allows you to do this, but it only allows positive values of t. So, use the command = TDIST( jtj , degrees of freedom, number of tails) = TDIST(3.27, 24, 2) = 0.00324 Thus, we can reject the null hypothesis at the 5% signi…cance level. In fact, we are 99.7% sure that the null hypothesis is false. Maple allows for both positive and negative values of t. Using the table above, the p-value can be found using the formula 2 [1 TDist( jtj , degrees of freedom)] TDist(3:27; 24) = 0:99838: The p-value is 2(1 0:99838) = 0:00324, which is the same answer we got from Excel. Now test the null hypothesis that = 60. This time the test statistic is t = 0:663 which yields a p-value of 0.514. We cannot reject this null hypothesis. What about the hypothesis that 65? The sample mean is 58.73, which is less than 65, so we should still do the test. (If the sample mean had been above 65, there is no way we could reject the hypothesis.) This is a one-tailed test based on the same test statistic which we have already CHAPTER 17. HYPOTHESIS TESTING 208 computed, t = 3:27. We have to change the Excel command to reduce the number of tails: =TDIST(3.27, 24, 1) = 0.00162 Once again we reject the hypothesis at the 5% level. Notice, however, that the p-value is twice what it was for the two-tailed test. This is as it should be, as you can …gure out by looking at the above table. 17.3.2 Example 2 You draw a sample of 100 numbers drawn from a normal distribution with mean and variance 2 . You compute the sample mean and sample variance, and they are x = 10 and s2 = 16. The null hypothesis is H0 : =9 Do the data support or reject the hypothesis? Compute the t-statistic t= x 10 9 p =p p = 2:5 s= n 16= 100 This by itself does not tell us anything. We must plug it into the appropriate t distribution. The t-statistic has 100 1 = 99 degrees of freedom, and we can …nd TDist(2:5; 99) = 0:992 97 and we reject if this number is either less than 0.025 or greater than 0.975. It is greater than 0.975, so we can reject the hypothesis. Another way to see it is by computing the p-value p = 2(1 TDist(2:5; 99)) = 0:014 which is much smaller than the 0.05 required for rejection. 17.3.3 Example 3 Use the same information as example 2, but instead test the null hypothesis H0 : = 10:4: 209 CHAPTER 17. HYPOTHESIS TESTING Do the data support or reject this hypothesis? Compute the t-statistic t= 10 10:4 x p =p p = s= n 16= 100 1:0 We can …nd TDist( 1; 99) = 0:16 :and we reject if this number is either less than 0.025 or greater than 0.975. It is not, so we cannot reject the hypothesis. Another way to see it is by computing the p-value p = 2(1 TDist(1; 99)) = 0:32 which is much larger than the 0.05 required for rejection. 17.4 Problems 1. Consider the following random sample from a normal distribution with mean and standard deviation 2 : 134; 99; 21; 38; 98; 19; 53; 52; 115; 30; 65; 149; 4; 55; 43; 26; 122; 47; 54; 97; 87; 34; 114; 44; 26; 98; 38; 24; 30; 86: (a) Test the hypothesis that hypothesis? = 0. Do the data support or reject the (b) Test the hypothesis that = 30. (c) Test the hypothesis that 65. (d) Test the hypothesis that 100. 2. Answer the following questions based on this random sample generated from a normal distribution with mean and variance 2 . 89; 51; 12; 17; 71 39; 47; 37; 42; 75 78; 67; 20; 9; 9 44; 71; 32; 13; 61 210 CHAPTER 17. HYPOTHESIS TESTING (a) What are the best estimates of and 2 ? (b) Test the hypothesis that the hypothesis? = 40. Do the data support or reject (c) Test the hypothesis that the hypothesis? = 60. Do the data support or reject CHAPTER 18 Solutions to end-of-chapter problems Solutions for Chapter 2 1. (a) f 0 (x) = 72x2 (x3 + 1) + (c) f 0 (x) = e2x (d) f 0 (x) = 9 (e) f (x) = 14x3 x1: 3 c (d cx)2 (c) h0 (x) = 1 4x3 20 x5 2) ln x ax2 ) (b 2a d xcx 24 1 2x3 4(6x (d) f 0 (x) = (1 (42x2 2: 7 x1: 3 2. (a) f 0 (x) = 24x (b) g 0 (x) = + 20 (4x 2)6 (b) f 0 (x) = 0 3 x x)e 2 ln 3x)=(4x3 ) ln 3x = (1 2)=(3x2 2x + 1)5 x 211 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 212 (e) g 0 (x) = 9 (9x 2x2 8)2 3 p 5x3 + 6 4 x 9x 8 p 5x3 + 6 3. Let f (x) = x2 . Then f (x + h) f (x) h!0 h 2 (x + h) x2 lim h!0 h x2 + 2xh + h2 x2 lim h!0 h 2 2xh + h lim h!0 h lim (2x + h) f 0 (x) = lim = = = = h!0 = 2x: 4. Use the formula f (x + h) h!0 h f (x) lim = lim h!0 = lim 1 x+h h h!0 x+h x(x+h) x x(x+h) h h!0 = lim 1 x h x(x+h) h h h!0 xh(x + h) 1 = lim = h!0 x(x + h) = lim 5. (a) Compute f 0 (3) = 18 < 0, so it is decreasing. (b) Compute f 0 (13) = 1 13 (c) Compute f 0 (4) = 5:0e (d) Compute f 0 (2) = 9 16 > 0, so it is increasing. 4 < 0, so it is decreasing. > 0, so it is increasing. 1 x2 2x2 3 15 x2 p 2 9x 8 5x3 + 6 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 213 6. (a) f 0 (x) = increasing. 3 x2 +4x (b) f 0 (x) = 1 x ln2 x 0 1) = 2) (x2x+4 2 +4x)2 and f ( (3x and f 0 (e) = 1 e 1 9 > 0. It is < 0. It is decreasing. (c) f 0 (x) = 10x + 16 and f 0 ( 6) = 44 < 0. It is decreasing. 7. (a) The FOC is f 0 (x) = 10 8x = 0 so x = 5=4. Compute f 00 (5=4) = 8 < 0, so this is a maximum. 6 = 0 so x = 141=0:3 . Compute (b) The FOC is f 0 (x) = x84 0:3 f 00 (141=0:3 ) = 2: 721 7 10 4 < 0, so this is a maximum. (c) The FOC is f 0 (x) = 4 x3 = 0 so x = 3=4. Compute f 00 (3=4) = 16 > 0, so this is a minimum. 3 8. (a) The foc is f 0 (x) = 8x 24 = 0, which is solved when x = 3. Also, f 00 (x) = 8 > 0, so it is a minimum. (b) The foc is f 0 (x) = 20=x 4 = 0, which is solved when x = 5. Also, f 00 (x) = 20=x2 < 0, so it is a maximum. (c) The foc is f 0 (x) = 1 + 6 = 0; x+2 x+1 (x + 2)2 and x = 11 . The second which has two solutions: x = 13 6 6 x+1 13 2 00 00 derivative is f (x) = (x+2)2 2 (x+2)3 , and f ( 6 ) = 432 while f 00 ( 11 ) = 432. The function has a local maximum when x = 13 6 6 . and a local minimum when x = 11 6 9. (a) Compute f 00 (x) = 2a. Need a < 0: (b) Need a > 0. 10. (a) The problem is max b(m) m c(m) The FOC is b0 (m) = c0 (m) Interpretation is that marginal bene…t equals marginal cost. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 214 (b) We need b00 (m) c00 (m) 0. A better way is to assume b00 (m) 0 and c00 (m) 0. This is diminishing marginal bene…t and increasing marginal cost. (c) The problem is max wm c(m) m The FOC is w = c0 (m) Interpretation is that marginal e¤ort cost equals the wage. (d) c00 (m) 0, or increasing marginal e¤ort cost. 11. The …rst-order condition is 0 90 (L) = p L 90 = 0: Solving for L gives us L = 1. Bilco devotes 1 unit of labor to widget production and the other 59 to gookey production. It produces W = 20(1)1=2 = 20 widgets and G = 30(59) = 1770 gookeys. Solutions for Chapter 3 1. (a) (26; 3; 33; 25) (b) x y and x < y and x y (c) 77 (d) Yes. x x = 65, y y = 135, and (x + y) (x + y) = 354. We have p p p p x x + y y = 65 + 135 = 19:681 and p (x + y) (x + y) = p 354 = 18: 815: 2. (a) (18; 8; 48; 20) (b) 7 p p p p p = 5: 099, and (x + y) (x + y) = (c) px x = 65 = 8: 062 3, y y = 26 p p 77 = 8: 775 0, which is smaller than 65 + 26 = 13: 161. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 215 3. (a) fx (x; y) = 8x (b) fy (x; y) = 6y (c) 12y + 18 12x 9 9 ; 8 4 4. (a) fx (x; y) = 16y (b) fy (x; y) = 16x 4: 2=y 2 . (c) The two foc’s are 16y 4 = 0 and 16x 2=y 2 = 0. The …rst one implies that y = 14 . Plugging this into the second expressions yields 16x 2 = 0 y2 2 16x = = 32 ( 41 )2 x = 2 The critical point is (x; y) = (2; 14 ). 5. (a) 3 ln x + 2 ln y = k (b) Implicitly di¤erentiate 3 ln x + 2 ln y(x) = k with respect to x to get 3 2 dy + =0 x y dx dy = dx 6. (a) (q) = p(q)q cq = 120q 3=x = 2=y 4q 2 3y 2x cq (b) The FOC is 120 8q c = 0 q = 15 (c) Using the answer to (b), we have dq =dc = c 8 1=8 < 0 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 216 c 8 (d) Plug q = 15 into (q) to get c 8 (q) = 120 15 = 1800 15c = 900 15c + c 8 4 15 2 c 15 c2 16 900 + 15c 15c + c 8 c2 8 c2 16 (e) Di¤erentiating yields 0 (c) = 15 + c 8 (f) Compare the answers to (b) and (e). Note that q is also the partial derivative of (q) = p(q)q cq with respect to c, which is why this works. 7. (a) Implicitly di¤erentiate to get 30x dx dx + 3a + 3x da da 5 dx 5x = 0: + a da a2 Solving for dx=da yields 30x + 3a 5 a dx = da dx = da 5x a2 3x + 5x a2 30x + 3a 3x + 5 a = x 3a2 + 5 a 3a2 + 30xa (b) Implicitly di¤erentiate to get 12xa dx + 6x2 = 5 da 5a2 dx da 10xa: Solving for dx=da yields 12xa + 5a2 dx = 5 10xa 6x2 da dx 5 10xa 6x2 = da 12xa + 5a2 5 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 217 8. (a) The foc is 30 p L w=0 and solving it for L gets p 30 = w L p 30 L = w 900 L = w2 (b) Since L = 900 w2 we have dL 1800 <0 = dw w3 The …rm uses fewer workers when the wage rises. (c) Plugging L into the pro…t function yields r 900 900 900 w = = 30 4 2 2 w w w and from there we …nd d = dw 900 < 0: w2 Pro…t falls when the wage rises. This happens for two reasons. One is that the …rm must pay workers more, and the other is that it uses fewer workers (see part b) and produces less output. 9. (a) Implicitly di¤erentiate to …nd dK=dL: FK (K; L) dK + FL (K; L) = 0 dL dK FL (K; L) = dL FK (K; L) (b) Both FL and FK are positive, so dK=dL = isoquant slopes downward. FL =FK < 0 and the CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 218 Solutions for Chapter 4 1. Set up the Lagrangian L(x; y; ) = 12x2 y 4 + (120 2x 4y): The foc’s are @L = 24xy 4 2 = 0 @x @L = 48x2 y 3 4 = 0 @y @L = 120 2x 4y = 0 @ This is three equations in three unknowns, so now we solve for the values of x, y, and . There are many ways to do this, and one of them can be found on page 39. Here is another. Solve the third equation for x: 120 2x 4y = 0 x = 60 2y Substitute this into the …rst two equations 24xy 4 48x2 y 3 2 4 = 0 = 0 to get 24(60 2y)y 4 48(60 2y)2 y 3 Multiply the top equation by equation to get 48(60 The terms with 2y)2 y 3 4 2 4 = 0 = 0 2 and add the result to the second 48(60 2y)y 4 4 =0 in them cancel out, and we are left with 48(60 2y)2 y 3 48(60 2y)y 4 = 0 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 219 Divide both sides by 48(60 (60 2y)y 3 to get 2y) y = 0 60 3y = 0 y = 20 Substitute this back into things we know to get x = 60 2y = 20 and = 12(60 2y)y 4 = 12(20)(204 ) = 38; 400; 000: 2. The Lagrangian is L(a; b; ) = 3 ln a + 2 ln b + (400 12a 14b) The FOCs are @L 3 = 12 = 0 @a a @L 2 = 14 = 0 @b b @L = 400 12a 14b = 0 @ Solving the …rst two yields a = 3=12 and b = 2=14 . into the third equation gives us 400 12 3 12 14 400 2 = 0 14 3 2 = 0 400 = = 5 1 5 = 400 80 Plugging into the earlier expressions, 3 3 240 a= = = = 20 12 12=80 12 and b= 2 2 160 80 = = = : 14 14=80 14 7 Substituting CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 220 3. The Lagrangian is L(x; y; ) = 16x + y + (1 x1=4 y 3=4 ): The FOCs are x1=4 y 3=4 y 3=4 1 @L = 16 = 16 =0 @x 4 x 4 x 1=4 @L 3 x 3 x1=4 y 3=4 = 1 =0 =1 @y 4 y 4 y @L = 1 x1=4 y 3=4 = 0 @ From the third FOC we know that x1=4 y 3=4 = 1; so the other two FOCs simplify to = 64x and 4 = y: 3 Setting these equal to each other gives us 4 y = 64x 3 y = 48x: Plugging this into the third FOC yields x1=4 y 3=4 = 1 x1=4 (48x)3=4 = 1 x = 31=4 1 = : 483=4 24 We can then solve for y = 48x = 2 31=4 and = 4y 8 31=4 = : 3 3 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 221 4. Set up the Lagrangian L(x; y; ) = 3xy + 4x + (80 4x 12y): The foc’s are @L = 3y + 4 4 = 0 @x @L = 3x 12 = 0 @y @L = 80 4x 12y = 0 @ The …rst equation reduces to y = 4( 1)=3 and the second equation tells us that x = 4 . Substituting these into the third equation yields 80 4(4 ) 80 4x 12y = 12(4)( 1)=3 = 96 32 = = 0 0 0 3 Plugging this into the equations we already derived gives us the rest of the solution: x = 4 = 12 y = 4( 1)=3 = 8=3: 5. Set up the Lagrangian L(x; y; ) = 5x + 2y + (80 3x The foc’s are @L = 5 3 2y = 0 @x @L = 2 2x = 0 @y @L = 80 3x 2xy = 0 @ 2xy) CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 222 5 3 2y = 0 2x = 0 2xy = 0 2 3x 80 Now we solve these equations. The third one reduces to 80 3x 2xy = 0 2xy = 80 80 y = 3x 3x 2x and the second one reduces to 2 2x = 0 1 = : x Substitute these into the …rst one to get 5 3 1 x 2 5 3 80 3x 2x 2y 1 x = 0 = 0 Multiplying through by x2 yields 5x2 3x 80 + 3x 5x2 x2 x = 0 = 80 = 16 = 4 Note that we only use the positive root in economics, so x = 4. Substituting into the other two equations yields y= 80 3x 2x and = = 1 1 = : x 4 17 2 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 223 6. (a) 0 (x) = 400 + 4x > 0. It says that increasing the size of the farm leads to increased pro…t, which seems sensible when the farm starts o¤ small. (b) 00 (x) = 4. This is questionable. But, it could arise because of increasing returns to scale or because of …xed inputs. (c) The Lagrangian is L(x; ) = 400x + 2x2 + (10 x): The FOCs are @L = 400 + 4x @x @L = 10 x = 0 @ =0 The second one tells us that x = 10 and the …rst one tells us that = 400 + 4x = 440: (d) It is the marginal value of land. (e) That would be 0 (10) = 440. This, by the way, is why the lame problem is useful. Note that the answers to (d) and (e) are the same. (f) No. remember that 0 (x) > 0, so is increasing and more land is better. Pro…t is maximized subject to the constraint. Obviously, constrained optimization will require a di¤erent set of second order conditions than unconstrained optimization does. p 7. (a) 0 (L) = 30= L 10 which is positive when L < 9. We would hope for an upward-sloping pro…t function, so this works, especially since L is only equal to 4. (b) 00 (L) = 15=L3=2 which is negative. Pro…t grows at a decreasing rate, which makes sense. (c) The Lagrangian is p L(L; ) = 30 4L 10L + (4 L) CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 224 The foc’s are @L 30 10 =0 = p @L L @L = 4 L=0 @ The second equation can be solved to get L = 4. Plugging L = 4 into the …rst equation yields 30 p 10 = 0 L 30 p 10 = 0 L 5 = (d) The Lagrange multiplier is always the marginal value of relaxing the constraint, where the value comes from whatever the objective function measures. In this case the objective function is the pro…t function, and the constraint is on the number of workers the …rm can use at one time, so the Lagrange multiplier measures the marginal pro…t from adding workers. (e) This is 0 p (4) = 30= 4 10 = 5: Note that this matches the answer from (c). (f) No. The …rst derivative of the pro…t function is positive (and equal to 5) when L = 4, which means that pro…t is increasing when L is 4. The second derivative does not tell us whether we are at a maximum or minimum when there is a constraint. 8. (a) The Lagrangian is L(x; y; ) = x y 1 + (M px x py y): The FOCs are @L = x @x @L = (1 @y @L = M @ 1 1 y )x y px x px = 0 py = 0 py y = 0: CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 225 Rearrange the …rst two to get y 1 x = px (1 x y ) = py : Set them equal to each other to get y 1 x y x 1 (1 = px y x (1 = y (1 = x (1 y = ) x y py ) px py ) px py ) px x: py Now substitute this into the budget constraint to get px x + py y = M (1 ) px x = M px x + py py (1 ) px x = M px x + px x = M 1 + (1 M : x = px ) = M Substituting this back into what we found for y yields y = = = (1 (1 (1 py ) px x py ) px M py px )M : CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 226 (b) These are easy. @x @M @y @M = >0 px 1 = > 0: py (c) Again, these are easy. M @x <0 = @px p2x @y = 0: @px The demand curve for good x is downward-sloping, and it is independent of the price of the other good. 9. (a) Denote labor devoted to widget production by w and labor devoted to gookey production by g. The Lagrangian is L(w; g; ) = (9)(20w1=2 ) + (3)(30g) (11)(w + g) + (60 w g): The foc’s are 1 @L = 90w 2 11 =0 @w @L = 90 11 =0 @g @L = 60 w g = 0 @ The second equation says that equation yields 90 p w 11 = 79. Plugging this into the …rst 79 = 0 p 90 = 90 w w = 1 The third equation then implies that g = 60 w = 59. These are the same as the answers to the question 4 on the …rst homework. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 227 (b) The Lagrange multiplier is the marginal value of adding workers. 10. (a) The farmer’s problem is max LW L;W s.t. 2L + 2W = F W = S (b) The Lagrangian is L(L; W; ; ) = LW + (F 2L 2W ) + (S W ): The …rst-order conditions are @L @L @L @W @L @ @L @ = W 2 =0 = L 2 =0 = F 2L = S W =0 2W = 0 We must solve this set of equations: W = S (from fourth equation) L = F=2 S (from third equation) = S=2 (from second equation) = F=2 2S (from …rst equation) (c) It depends. The marginal impact on area comes directly from the Lagrange multipliers. is the marginal impact of having a longer fence while keeping the shortest side …xed, and is the marginal impact of lengthening the shortest side while keeping the total CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 228 fence length constant. We want to know which is greater, S=2 or F=2 2S. We can …nd S=2 5S=2 S F=2 2S F=2 F=5: When the shortest side is more than one-…fth of the total amount of fencing, the farmer would rather lengthen the fence than lengthen the shortest side. When the shortest side is smaller than a …fth of the fence lenght, she would rather lengthen that side, keeping the total fence length …xed. Solutions for Chapter 5 1. (a) The solution to the alternative problem is (x; y) = (8; 38 ). Note that 4 8 + 83 = 34 23 > 20, so the second constraint does not hold. (b) The solution to the alternative problem is (x; y) = ( 10 ; 20 ). Note 3 3 20 80 10 that 2 3 + 3 3 = 3 > 24, so the …rst constraint does not hold. (c) If the solution to the alternative problem in (a) had satis…ed the second constraint, the second constraint would have been nonbinding and its Lagrange multiplier would have been zero. This is not what happened, though, so the second constraint must bind, in which case 2 > 0. Similarly, part (b) shows us that the …rst constraint must also bind, and so 1 > 0. (d) Because both constraints bind, the problem becomes maxx;y x2 y s.t. 2x + 3y = 24 4x + y = 20 This is easy to solve because there is only one point that satis; 28 ). Now …nd the Lagrange …es both constraints: (x; y) = ( 18 5 5 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 229 multipliers. The FOCs for the equality-constrained problem are 2xy 2 1 4 2 x2 3 1 2 24 2x 3y 20 4x y = = = = 0 0 0 0 We already used the last two to …nd x and y. Plug those values into the …rst two to get two equations in two unknowns: 2 4 1 3 1 The solution to this is ( 1 ; + 2) 2 2 1008 25 324 = 25 = 144 1188 = ( 125 ; 125 ). 2. (a) The solution to the alternative problem is (x; y) = (8; 38 ). Note that 4 8 + 83 = 34 32 < 36, so the second constraint does hold this time. (b) The solution to the alternative problem is (x; y) = (6; 12). Note that 2 6 + 3 12 = 48 > 24, so the …rst constraint does not hold. (c) The solution to the alternative problem in (a) satis…es the second constraint, so the second contrainst is nonbinding. Therefore 1 > 0 and 2 = 0. (d) Because only the …rst constraint binds, the problem becomes maxx;y x2 y s.t. 2x + 3y = 24 We know from part (a) that (x; y) = (8; 38 ). We also know that To …nd 1 use the FOCs for the equality-constrained 2 = 0. problem: 2xy 2 1 = 0 x2 3 1 = 0 24 2x 3y = 0 Plug x = 8 into the second equation to get 1 = 64 . Or, plug 3 8 x = 8 and y = 3 into the …rst equation to get the same thing. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 230 3. (a) Setting conditions 2 = 0 in the original Lagrangian we get the …rst-order @L = 4y @x @L = 4x @y @L = 36 @ 1 We solve these for x, y, and 1 6x 1 4 1 x (b) Setting 1 45 2 + =0 4y = 0 and get 9 63 x= ,y= , 2 8 We then have 5x + 2y = is satis…ed. =0 63 4 = 153 4 1 = 9 2 < 45 and the second constraint = 0 in the original Lagrangian, the foc’s are @L = 4y @x @L = 4x @y @L = 45 @ 2 6x 2 5 2 5x 2 =0 =0 2y = 0 The solution is 45 180 90 ,y= ; 2= 13 13 13 720 765 45 We then have x + 4y = 13 + 13 = 13 > 36 and the …rst constraint is not satis…ed. x= (c) Part (a) shows that we can get a solution when the …rst constraint binds and the second doesn’t, and part (b) shows that we cannot get a solution when the second constraint binds but the …rst does not. So, the answer comes from part (a), with 9 63 x= ,y= , 2 8 1 9 = , 2 2 = 0: CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 231 4. (a) The foc’s are @L = 3y @x @L = 3x @y @L = 24 @ 1 The solution is 8 4 1 1 x =0 =0 4y = 0 13 20 ,y= , 1=5 3 3 26 + 3 = 42 > 30 the second constraint is not x= and since 5x + 2y = satis…ed. 100 3 (b) The foc’s are @L = 3y @x @L = 3x @y @L = 30 @ 2 8 2 5x 5 2 2 =0 =0 2y = 0 3y 8 5 2 = 0 3x 2 2 = 0 30 5x 2y = 0 The solution is 53 37 37 ,y= , 2= 15 6 10 > 24 the …rst constraint is not satis…ed. x= and since x + 4y = 189 5 (c) Since when one constraint binds the other fails, they must both bind. The place where they both bind is the intersection of the two "budget lines," or where the following system is solved: x + 4y = 24 5x + 2y = 30 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 232 The solution is x = 4, y = 5. Now we have to …nd the values of 1 and 2 . To do this, go back to the foc’s for the entire original Lagrangian: @L = 3y 8 5 2=0 1 @x @L = 3x 4 1 2 2 = 0 @y @L = 24 x 4y = 0 @ 1 @L = 30 5x 2y = 0 @ 2 Plug the values for x and y into the …rst two equations to get 15 8 12 and solve for 1 and 2. 5 2 1 4 1 2 2 = 0 = 0 The solution is 1 = 23 9 and 2 = 98 . 5. (a) K(x; y; ) = x2 y + [42 4x 2y]: (b) @K @x @K y @y @K @ x = x(2xy 4 )=0 = y(x2 2 )=0 = 4x (42 x; y; 2y) = 0 0 (c) First notice that the objective function is x2 y, which is zero if either x or y is zero. Consequently, neither x 0 nor y 0 can be binding. The other, budget-like constraint is binding because x2 y is increasing in both arguments, and so > 0: The KuhnTucker conditions reduce to 2xy x2 42 4x 4 = 0 2 = 0 2y = 0 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 233 Solving yields (x; y; ) = (7; 7; 49 ). 2 6. (a) K(x; y; ) = xy + 40x + 60y + (12 x y) (b) @K @x @K y @y @K @ x = x(y + 40 )=0 = y(x + 60 )=0 = (12 x; y; x y) = 0 0 (c) This one is complicated, because we can identify three potential solutions: (i) x is zero and y is positive, (ii) x is positive and y is zero, and (iii) both x and y are positive. The only thing to do is try them one at a time. Case (i): x = 0. Then y = 12 from the third equation, and = 60 from the second equation. The value of the objective function is xy + 40x + 60y = 720. Case (ii): y = 0. Then x = 12 from the third equation, and = 40 from the …rst equation. The value of the objective function is 480. This case is not as good as case (i), so it cannot be the answer. Case (iii): x; y > 0. Divide both sides of the …rst K-T condition by x, which is legal since x > 0, divide the second by y, and divide the third by . We get y + 40 x + 60 12 x = 0 = 0 y = 0 The solution to this system of equations is x = This is not allowed, though, because x < 0. The …nal solution is case (i): x = 0, y = 12, 4, y = 16, = 60. = 56. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 234 Solutions for Chapter 6 19 2 1. (a) 16 3 10 0 13 18 (b) 0 7 @ 19 (c) 23 (d) 39 0 14 @ 21 2. (a) 9 22 61 1 5 1 1 2 2 2 A 36 1 1 23 32 A 11 1 7 0 (b) @ 12 14 A 0 9 0 33 7 (c) 13 32 11 36 (d) 84 3. (a) 14 (b) 134 4. (a) 21 (b) 12 5. In matrix form the system of equations is 1 10 1 0 0 1 x 6 2 3 @ 2 4 1 A@ y A = @ 2 A: 8 z 3 0 1 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 235 Using Cramer’s rule we get 0 1 1 2 3 1 A det @ 2 4 8 0 1 80 1 = 0 x= = 40 2 6 2 3 1 A det @ 2 4 3 0 1 1 0 6 1 3 2 1 A det @ 2 3 8 1 97 1= 0 y= 2 6 2 3 A @ 1 det 2 4 3 0 1 1 0 6 2 1 2 A det @ 2 4 3 0 8 224 1= 0 z= = 112 2 6 2 3 1 A det @ 2 4 3 0 1 6. In matrix form the system 0 5 2 @ 3 1 0 3 of equations is 1 10 1 0 9 x 1 0 A@ y A = @ 9 A: 15 z 2 Using Cramer’s rule we get 0 9 2 @ 9 1 det 15 3 0 x= 5 2 @ 1 det 3 0 3 1 1 0 A 2 60 1 = 11 1 A 0 2 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 236 0 7. (a) 1 2 1 8 3 2 0 1 (b) 21 @ 1 1 8. (a) 4 2 1 14 5 @ det 3 0 0 y= 5 det @ 3 0 0 5 @ det 3 0 0 z= 5 @ det 3 0 2 2 0 9 9 15 2 1 3 2 1 3 2 1 3 1 1 0 A 2 81 1= 11 1 0 A 2 1 9 9 A 15 39 1 = 11 1 A 0 2 1 1 1 A 1 1 4 1 5 4 3 1 @ 0 15 5 A (b) 25 0 5 10 0 Solutions for Chapter 7 1. (a) The determinant of the matrix 0 3 6 @ 2 0 1 1 is 1 0 5 A 1 33, and so there is a unique solution. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 237 (b) The determinant of the matrix 1 0 4 1 8 @ 17 8 10 A 3 2 2 is 0, and so there is not a unique solution. To …nd out whether there is no solution or an in…nite number, get the augmented matrix in row-echelon form. 1 0 160 4 1 8 @ 17 8 10 200 A 40 3 2 2 0 1 4 1 8 160 15 @ 0 24 480 A 4 5 0 4 8 160 1 0 4 1 8 160 15 @ 0 24 480 A 4 0 0 0 0 Since the bottom row is zeros all the way across, there are in…nitely many solutions. (c) The determinant of the matrix 1 0 2 3 0 @ 3 0 5 A 2 6 10 is 0, and so there is not a unique solution. To …nd out whether there is no solution or an in…nite number, get the augmented matrix in row-echelon form. 1 0 2 3 0 6 @ 3 0 5 15 A 2 6 10 18 Multiply the …rst row by 2 and add it to the third row: 0 1 2 3 0 6 @ 3 0 5 15 A 6 0 10 30 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 238 Multiply the second row by 2 and subtract it from the third row: 0 1 2 3 0 6 @ 3 0 5 15 A 0 0 0 0 There is a row of all zeros, so there is an in…nite number of solutions. (d) The determinant of the matrix 0 4 1 @ 3 0 5 1 1 8 2 A 2 is 0, and so there is not a unique solution. To …nd out whether there is no solution or an in…nite number, get the augmented matrix in row-echelon form. 1 0 4 1 8 30 @ 3 0 2 20 A 5 1 2 40 Add the top row to the bottom 0 4 1 @ 3 0 9 0 Multiply the middle row by row: 0 4 @ 3 0 row: 8 2 6 1 30 20 A 70 3 and subtract it from the bottom 1 1 8 30 0 2 20 A 0 0 10 Since the bottom row is zeros all the way across except for the last column, there is no solution. (e) The determinant of the matrix 0 6 1 @ 5 2 0 1 is 27, there is a unique solution. 1 1 2 A 2 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 239 2. (a) There is no inverse if the determinant is zero, which leads to the equation 6a + 2 = 0 1 : 3 a = (b) Setting the determinant equal to zero and solving for a yields 5a 5 = 0 a = 1 (c) There is no inverse if the determinant is zero, which leads to the equation 5a + 9 = 0 a = 9 : 5 (d) Setting the determinant equal to zero and solving for a yields 20a 35 = 0 7 a = 4 Solutions for Chapter 8 1. (a) Rewrite the system as Y = c((1 t)Y ) + i(R) + G M = P m(Y; R) Implicitly di¤erentiate with respect to t to get dY dR + i0 dt dt dY dR 0 = P mY + P mR dt dt dY dt = c0 Y + (1 t)c0 Write in matrix form: 1 (1 t)c0 i0 mY mR dY dt dR dt = c0 Y 0 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 240 Use Cramer’s rule to get: dY dt = c0 Y i0 0 mR 0 (1 t)c i0 mY mR = c0 Y (1 (1 mR t)c0 )mR + mY i0 which is c0 Y times the derivative from the lecture. It is negative, so an increase in the tax rate reduces GDP. (1 t)c0 c0 Y mY 0 0 0 (1 t)c i mY mR mY c0 Y (1 (1 t)c0 )mR + mY i0 1 dR = dt = which is c0 Y times the derivative from the lecture. It is negative, so an increase in the tax rate reduces the interest rate. (b) Implicitly di¤erentiate the system with respect to M to get dY dM dY dR + i0 dM dM dR dY + P mR 1 = P mY dM dM = (1 t)c0 Write in matrix form: (1 t)c0 mY i0 mR dY dM dR dM = 0 1 Use Cramer’s rule to get: dY dM 0 i0 1 mR = (1 t)c0 i0 mY mR i0 = (1 (1 t)c0 )mR + mY i0 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 241 Both the numerator and denominator are negative, making the derivative positive, and so an increase in money supply increases GDP. t)c0 0 mY 1 dR = 0 dt (1 t)c i0 mY mR (1 t)c0 = (1 (1 t)c0 )mR + mY i0 (1 The numerator is positive, making the derivative negative, and so an increase in money supply reduces the interest rate. (c) Implicitly di¤erentiate the system with respect to P to get dY dP dY dR + i0 dP dP dY dR 0 = m + P mY + P mR dP dP = (1 t)c0 Write in matrix form: 1 (1 t)c0 i0 mY mR dY dP dR dP = 0 m The derivatives are m times the derivatives from part (b), and so an increase in the price level reduces GDP and increases the interest rate. 2. (a) First simplify to two equations: Y = c(Y T ) + i(R) + G + x(Y; R) M = P m(Y; R) Implicitly di¤erentiate with respect to G to get dY dY dR dY dR = c0 + i0 + 1 + xY + xR dG dG dG dG dG dY dR 0 = P mY + P mR dG dG CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 242 Rearrange as dY dG c0 dY dG i0 dR dG dY dR xR = 1 dG dG dY dR mY + mR = 0 dG dG xY We can write this in matrix form c0 xY mY 1 i0 x R mR dY dG dR dG = 1 0 Now use Cramer’s rule to solve for dY =dG and dR=dG: i0 x R mR dY = 0 dG 1 c xY i0 x R mY mR mR = (1 c0 xY )mR + mY (i0 + xR ) 1 0 The numerator is negative. The denominator is negative. So, dY =dG > 0. c0 xY 1 mY 0 dR = 0 0 dG 1 c xY i xR mY mR mY = (1 c0 xY )mR + mY (i0 + xR ) 1 The numerator is negative and so is the denominator. Thus, dR=dG > 0. An increase in government spending increases both GDP and interest rates in the short run. (b) In matrix form we get 1 c0 xY mY i0 x R mR dY dG dR dG = c0 0 Thus, the derivatives are c0 times those in part (a), so an increase in tax revenue reduces both GDP and interest rates. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 243 (c) dY dM i0 xR mR = 0 1 c xY i0 x R mY mR 0 i + xR = (1 c0 xY )mR + mY (i0 + xR ) 1 0 Both the numerator and denominator are negative, making the derivative positive, and so an increase in money supply increases GDP. c0 xY 0 mY 1 = (1 t)c0 i0 mY mR 1 c0 xY = (1 c0 xY )mR + mY (i0 + xR ) 1 dR dM The numerator is positive, making the derivative negative, and so an increase in money supply reduces the interest rate. 3. (a) Rewrite the system as Y = c((1 t)Y ) + i(R) + x(Y; R) + G M = P m(Y; R) Y = Y Implicitly di¤erentiate with respect to G to get dY dR dY dR + i0 + xY + xR +1 dG dG dG dG dY dR dP + P mY + P mR 0 = m(Y; R) dG dG dG dY = 0 dG dY = (1 dG t)c0 The last line implies, obviously, that dY =dG = 0. This makes sense because Y is …xed at the exogenous level Y . Even so, let’s go through CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 244 the e¤ort of writing the equation in 0 1 (1 t)c0 xY i0 xR @ P mY P mR 1 0 Use Cramer’s rule to get dY = dG 1 matrix notation: 10 1 0 1 dY =dG 0 1 m A @ dR=dG A = @ 0 A dP=dG 0 0 1 i0 x R 0 0 P mR m 0 0 0 0 0 (1 t)c xY i xR 0 P mY P mR m 1 0 0 =0 where the result follows immediately from the row with all zeroes. The increase in government spending has no long-run impact on GDP. As for interest rates, t)c0 xY 1 0 P mY 0 m 1 0 0 0 0 (1 t)c xY i xR 0 P mY P mR m 1 0 0 1 dR = dG 1 (1 = m mi0 mxR = i0 1 > 0: + xR Increased government spending leads to an increase in interest rates in the long run. Finally, 1 dP = dG 1 t)c0 P mY 1 (1 t)c0 P mY 1 (1 xY xY i0 x R P mR 0 0 i xR P mR 0 1 0 0 0 m 0 = P mR > 0: mxR mi0 An increase in government spending leads to an increase in the price level. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 245 (b) This time implicitly di¤erentiate with respect to M to get dY dM dY dR dY dR + i0 + xY + xR dM dM dM dM dP dY dR + P mY + P mR 1 = m(Y; R) dM dM dM dY = 0 dM Write the equation in matrix notation: 1 0 1 10 0 0 dY =dM 1 (1 t)c0 xY i0 x R 0 @ P mY P mR m A @ dR=dM A = @ 1 A 0 dP=dM 1 0 0 = (1 t)c0 Use Cramer’s rule to get dY = dM 1 0 i0 x R 0 1 P mR m 0 0 0 (1 t)c0 xY i0 xR 0 P mY P mR m 1 0 0 =0 The increase in money supply has no long-run impact on GDP. As for interest rates, t)c0 xY 0 0 P mY 1 m 1 0 0 0 0 (1 t)c xY i xR 0 P mY P mR m 1 0 0 1 dR = dM 1 (1 = 0 mi0 mxR = 0: Increasing the money supply has no long-run impact on interest rates, either. Finally, 1 dP = dM 1 t)c0 P mY 1 (1 t)c0 P mY 1 (1 xY xY i0 xR P mR 0 0 i xR P mR 0 0 1 0 0 m 0 = i0 mi0 1 xR = > 0: mxR m CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 246 An increase the money supply leads to an increase in the price level. That’s the only long-run impact of an increase in money supply. 4. (a) Implicitly di¤erentiate the system with respect to I: dqD dp = Dp + DI dI dI dqS dp = Sp dI dI dqS dqD = dI dI Write it in matrix form: 1 1 0 10 0 DI dqD =dI 1 0 Dp @ 0 1 Sp A @ dqS =dI A = @ 0 A 0 dp=dI 1 1 0 Solve for dp=dI using Cramer’s rule: dp = dI 1 0 DI 0 1 0 1 1 0 1 0 Dp 0 1 Sp 1 1 0 = DI >0 DP SP where the result follows because DI > 0, Dp < 0, and Sp > 0. (b) Implicitly di¤erentiate the system with respect to w: dqD dp = Dp dw dw dp dqS = Sp + Sw dw dw dqD dqS = dw dw Write it in matrix form: 1 1 0 10 0 0 dqD =dw 1 0 Dp @ 0 1 Sp A @ dqS =dw A = @ Sw A 0 dp=dw 1 1 0 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 247 Solve for dp=dw using Cramer’s rule: dp = dw 1 0 0 0 1 Sw 1 1 0 1 0 Dp 0 1 Sp 1 1 0 = Sw DP SP > 0; where the result follows because Sw < 0, Dp < 0, and Sp > 0. 5. We can write the regression as y = X + e where 0 0 1 1 6 1 9 y = @ 2 A and X = @ 1 4 A : 5 1 3 The estimated coe¢cients are given by ^ = (X T X) 1 X T y = 1 62 146 23 : 6. (a) The matrix is XT X = 2 6 8 24 and its determinant is 0. 4 16 0 2 @ 6 4 1 8 24 A = 16 56 224 224 896 (b) The second column of x is a scalar multiple of the …rst, and so the two vectors span the same column space. The regression projects the y vector onto this column space, but there are in…nitely-many ways to write the resulting projection as a combination of the two column vectors. 7. ^ = (X T X) 1 X T y = 1 138 205 64 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 248 8. x3 = 12x2 , and so the new variable does not expand the space spanned by the columns of the data matrix. All it does is make the solution indeterminant, and the matrix X T X will not be invertible. To see this, note that if we add the column 1 0 1 0 1 2 24 4 14 168 B 1 3 36 C T C @ 14 54 648 A X=B @ 1 5 60 A and X X = 168 648 7776 1 4 48 The determinant of X T X is 0. Also, the third column is 12 times the second column. 9. (a) The eigenvalues are given by the solution to the problem 5 1 4 = 0: 2 Taking the determinant yields (5 )(2 ) 4 = 0 6 7 + 2 = 0 = 6; 1 Eigenvectors satisfy 5 4 When v1 v2 1 2 = 0 0 0 0 : : = 1, this is 4 1 4 1 v1 v2 = There are many solutions, but one of them is v1 v2 When 1 4 = : = 6, the equation is 1 4 1 4 v1 v2 = 0 0 : CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 249 Again there are many solutions, but one of them is v1 v2 = 1 1 : (b) Use the same steps as before. The eigenvalues are = 7 and = 6: When = 7 an eigenvector is (1; 3), and when = 6 an eigenvector is (1; 4). (c) The eigenvalues are = 7 and = 0. When = 7 an eigenvector is (3; 2) and when = 0 an eigenvector is (2; 1). (d) The eigenvalues are = 3 and = 2. When = 3 an eigenvector is (1; 0) and when = 2 an eigenvector is ( 4; 1). p p (e) The eigenvalues are = 2 + 2 p13 and = 2 2 13. When p = p 2 + 2 13 an eigenvector p is ( 2 13 8; 3) and when = 2 2 13 an eigenvector is (2 13 8; 3). 10. (a) Yes. The eigenvalues are are less than one. = 1=3 and = 1=5, both of which (b) No. The eigenvalues are = 5=4 and = 1=3. The …rst one is larger than 1, so the system is unstable. (c) Yes. The eigenvalues are = 1=4 and have magnitude less than one. = 2=3, both of which (d) No. The eigenvalues are = 2 and = 41=45. The …rst one is larger than 1, so the system is unstable. Solutions for Chapter 9 1. 1 2x23 + 3x22 8x1 A 6x1 x2 rf (x1 ; x2 ; x3 ) = @ 4x1 x3 1 0 28 rf (5; 2; 0) = @ 60 A 0 0 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 250 2. (a) The second-order Taylor approximation is f (x0 ) + f 0 (x0 )(x f 00 (x0 ) (x 2 x0 ) + x0 ) 2 We have f (1) f 0 (x) f 0 (1) f 00 (x) f 00 (1) = 2 = 6x2 = 11 = 12x = 12 5 and so the Taylor approximation at 1 is 2 11(x 1) 12 (x 2 1)2 = 6x2 + x + 7 (b) We have f (1) = 30 20 1 p + x x f 0 (x) = 10 f 0 (1) = f 00 (x) = 9 10 x f 00 (1) = 9 1 x2 3 2 and the Taylor approximation at 1 is 30 9(x 9 1) + (x 2 9 1)2 = x2 2 18x (c) We have f (x) = f 0 (x) = f 00 (x) = ex f (1) = f 0 (1) = f 00 (1) = e and the Taylor approximation is e + e(x e 1) + (x 2 1 1)2 = e x2 + 1 2 33 2 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 251 3. f (x) f (x0 ) + f 0 (x0 )(x = 12 1 x0 ) + f 00 (x0 )(x 2 x0 ) 2 1 x0 ) + f 00 (x0 )(x 2 x0 ) 2 4x2 2x 4. f (x) f (x0 ) + f 0 (x0 )(x = c + bx + ax2 The second-degree Taylor approximation gives you a second-order polynomial, and if you begin with a second-order polynomial you get a perfect approximation. 5. (a) Negative de…nite because a11 < 0 and a11 a22 a12 a21 = 7. (b) Positive semide…nite because a11 > 0 but a11 a22 a12 a21 = 0. (c) Inde…nite because a11 > 0 but a11 a22 25: a12 a21 = (d) Inde…nite because ja11 j > 0, 4 0 0 3 = 12, and 4 0 1 0 3 2 1 2 1 25: (e) Positive de…nite because jA1 j = 6 > 0 and jA2 j = 17 > 0: (f) Inde…nite because jA1 j = 4 < 0 but jA2 j = (g) Negative de…nite because jA1 j = 240 < 0: 2 < 0 and jA2 j = 7 > 0: (h) Inde…nite because jA1 j = 3 > 0, jA2 j = 8 > 0, and jA3 j = 44 < 0: 6. (a) Letting f (x; y) denote the objective function, the …rst partials are fx = y fy = 8y x and the matrix of second partials is fxx fxy fyx fyy = 0 1 1 8 : This matrix is inde…nite and so the second-order conditions for a minimum are not satis…ed. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 252 (b) Letting f (x; y) denote the objective function, the …rst partials are fx = 8 fy = 6 2x 2y and the matrix of second partials is fxx fxy fyx fyy 2 0 = 0 2 which is negative de…nite. The second-order conditions are satis…ed. (c) We have rf = and H= 5y 5x 0 5 4y 5 4 This matrix is inde…nite because jH1 j = 0 but jH2 j = The second-order condition is not satis…ed. 25 < 0: (d) We have rf = and H= 12x 6y 12 0 0 6 This matrix is positive de…nite because jH1 j = 12 > 0 and jH2 j = 72 > 0: The second-order condition is satis…ed. 7. If it is a convex combination there must be some number t 2 [0; 1] such that 6 1 11 : + (1 t) =t 0 4 2 Writing these out as two equations gives us 6 = 11t + ( 1)(1 t) CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 253 and 2 = 4t + 0(1 t): Solving the …rst one yields t = 7=12 and solving the second one yields t = 1=2. These are not the same so it is not a convex combination. 8. Let t be a scalar between 0 and 1. Given xa and xb , we want to show that f (txa + (1 t)xb ) tf (xa ) + (1 t)f (xb ) Looking at the left-hand side, f (txa + (1 t)xb ) = (txa + (1 = (xb + txa t)xb )2 txb )2 Looking at the right-hand side, tf (xa ) + (1 t)f (xb ) = tx2a + (1 = x2b + tx2a t)x2b tx2b Subtracting the left-hand side from the right-hand side gives us x2b + tx2a tx2b (xb + txa txb )2 = t(1 which has to be nonnegative because t, 1 all nonnegative. t)(xa xb ) 2 t, and anything squared are Solutions for Chapter 10 1. (a) x = 20 and y = 4. (b) There are two of them: x = 15 and y = 4, and x = 15 and y = 5. (c) Sum the probabilities along the row to get 0:35. (d) 0:03 + 0:17 + 0:00 + 0:05 + 0:04 + 0:20 = 0:49. (e) P (y 2jx 20) = P (y 0:23 23 2 and x 20) = = P (x 20) 0:79 79 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 254 (f) Bayes’ rule says P (y = 4jx = 20) = P (x = 20jy = 4) P (y = 4) : P (x = 20) We have P (y = 4jx = 20) = 0:20=0:44 = 5=11. Also, (0:20=0:27) (0:27) 20 5 P (x = 20jy = 4) P (y = 4) = = = : P (x = 20) 0:44 44 11 (g) Two events are statistically independent if the probability of their intersection equals the product of their probabilities. We have P (x 20) = 0:65 P (y 2 f1; 4g) = 0:42 P (x 20) P (y 2 f1; 4g) = (0:65)(0:42) = 0:273 P (x 20 and y 2 f1; 4g) = 0:24 They are not statistically independent. 2. (a) P (A) = 0:26 and P (B) = :18, so A is more likely. (b) The numbers in parentheses are (a; b) pairs: f(4; 1); (4; 3); (5; 3)g. (c) 0:73. (d) P (b = 2ja = 5) = P (b = 2 and a = 5)=P (a = 5) = 0:06=0:32 = 3=16 = 0:1875: (e) P (a and 3 and b 2 f1; 4g) = 0:45 P (b 2 f1; 4g) = 0:58 so 0:45 = 0:77586 0:58 (f) P (a 2 f1; 3g and b 2 f1; 2; 4g) = 0:14, but P (a 2 f1; 3g) = 0:36 and P (b 2 f1; 2; 4g) = 0:73. We have P (a 3jb 2 f1; 4g) = P ((a 2 f1; 3g)P (b 2 f1; 2; 4g) = 0:36 0:73 = 0:2628 6= 0:14: They are not statistically independent. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 255 3. We want P (disease j positive), which is P (disease j positive) = P (disease and positive) . P (positive) Note that P (disease and positive) = P (positive j disease) P (disease) 1 = 0:95 20; 000 = 0:0000475 and P (positive) = P (positive j disease) P (disease) + P (positive j healthy) P (healthy) 19; 999 1 + 0:05 = 0:95 20; 000 20; 000 = 0:0000475 + 0:0499975 = 0:050045 Now we get P (disease and positive) P (positive) 0:0000475 = 0:050045 = 0:000949 P (disease j positive) = In spite of the positive test, it is still very unlikely that Max has the disease. 4. Use Bayes’ rule: P (entrepreneur j old) = P (old j entrpreneur)P (entrepreneur) P (old) Your grad assistant told you that P (old j entrepreneur) = 0:8 and that P (entrepreneur) = 0:3. But she didn’t tell you P (old), so you must CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 256 calculate it: P (old) = P (old j doctor)P (doctor) +P (old j lawyer)P (lawyer) +P (old j entrpreneur)P (entrpreneur) = (0:6)(0:2) + (0:3)(0:5) + (0:8)(0:3) = 0:51 Plugging this into Bayes’ rule yields P (entrepreneur j old) = (0:8)(0:3) = 0:47 0:51 47% of old people are entrepreneurs. Solutions for Chapter 12 1. (a) Plugging in f (x) = 1=6 gives us Z 8 Z 1 8 xf (x)dx = xdx 6 2 2 8 1 2 = x 12 2 64 4 = =5 12 12 (b) Following the same strategy, Z Z 8 1 8 2 2 x dx x f (x)dx = 6 2 2 8 1 3 x = 18 2 8 512 = 28 = 18 18 2. Leibniz’ rule says Z b(t) Z d b(t) @f (x; t) f (x; t)dx = dx + b0 (t)f (b(t); t) dt a(t) @t a(t) a0 (t)f (a(t); t): CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 257 Here f (x; t) = tx2 , b(t) = t2 , and a(t) = t2 so @f (x; t) = x2 , f (b(t); t) = t (t2 )2 = t5 , and f (a(t); t) = t ( t2 )2 = t5 . @t Leibniz’ rule then becomes Z Z 2 d t 2 tx dx = dt t2 = t2 x2 dx + (2t)(t5 ) ( 2t)(t5 ) t2 x3 3 t2 + 2t6 + 2t6 t2 6 t t6 + 4t6 3 3 14 6 t: = 3 = 3. Use Leibniz’ rule: Z b(t) Z @f (x; t) d b(t) dx + b0 (t)f (b(t); t) a0 (t)f (a(t); t) f (x; t)dx = dt a(t) @t a(t) Z 4t2 Z 4t2 d tx3 dx + (8t) t2 (4t2 )3 ( 3) t2 ( 3t)3 t2 x3 dx = 2 dt 3t 3t = 2 4 tx 4 = 128t9 = 640t9 4t2 + 512t9 81t5 3t 81 5 t + 512t9 2 243 5 t 2 81t5 4. Let F (x) denote the distribution function for U (a; b), and let G(x) denote the distribution function for U (0; 1). Then 8 x<a < 0 x a for a x<b F (x) = : b a 1 x b CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 258 and 8 x<0 < 0 x for 0 x < 1 G(x) = : 1 x 1 First-order stochastic dominance requires F (x) ments on a and b are a b G(x). The require- 0 1 The easiest way to see this is by graphing it. equations, if 0 a 1 b we can write 8 0 > > > > x < x xb aa for G(x) F (x) = > > 1 xb aa > > : 0 But, from looking at the 0 a 1 x<0 x<a x<1 x<b x b Note that x x b bx a = a which is positive when b 1 x b which is positive when b ax x + a a(1 x) (b 1)x = + b a b a b a 1 a b = a x a a b 0, and x+a b = a b x: So G(x) x a F (x) as desired. Solutions for Chapter 13 1. (a) = (:10)(7)+(:23)(4)+(:40)(2)+(:15)( 2)+(:10)( 6)+(:02)( 14) = 1: 24 (b) 2 = (:10)(7 1: 24)2 + (:23)(4 1: 24)2 + (:40)(2 (:15)( 2 1: 24)2 + (:10)( 6 1: 24)2 + (:02)( 14 16: 762: 1: 24)2 + 1: 24)2 = CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 259 2. (a) The means are f = (10)(:15) + (15)(:5) + (20)(:05) + (30)(:1) + (100)(:2) = 33 g = (10)(:2) + (15)(:3) + (20)(:1) + (30)(:1) + (100)(:3) = 41: 5: and (b) The variances are 2 f = (10 33)2 (:15) + (15 33)2 (:5) + (20 +(30 33)2 (:1) + (100 = 1148:5 33)2 (:05) 33)2 (:2) and 2 g = (10 41:5)2 (:2) + (15 41:5)2 (:3) + (20 +(30 41:5)2 (:1) + (100 = 1495:3 41:5)2 (:3) (c) The standard deviations are p 1148:5 = 33: 890 f = and g = p 1495:3 = 38: 669 3. (a) F (x) = Z 0 = = Z x f (t)dt x 0 2 x t 0 2 = x: (b) All those things hold. 2tdt 41:5)2 (:1) CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 260 (c) = Z 1 x 2xdx 0 = 2 Z 1 x2 dx 0 1 2 3 = x 3 2 : = 3 0 (d) First …nd 2 E[~ x] = Z 1 x2 2xdx 0 Z = 2 1 x3 dx 0 2 4 x 4 1 = . 2 1 = 0 Then note that 2 = E[~ x2 ] 4. (a) For x 2 [0; 4] we have F (x) = 2 Z = 0 x 1 2 4 1 = : 9 18 1 tdt 8 x 1 2 t = 16 0 1 2 = x 16 Outside of this interval we have F (x) = 0 when x < 0 and F (x) = 1 when x > 1. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 261 (b) Using F (x) = x2 =16, we get F (0) = 0, F (4) = 1, and F 0 (x) = x=8 0. (c) = Z 4 1 8 x dx = 8 3 x 0 (d) 2 = Z 4 x 0 2 8 3 8 1 x dx = 8 9 5. The mean of the random variable x~ is a , where The variance is is the mean of x~. V ar(a~ x) = E[(a~ x a )2 ] = E[a2 (~ x )2 ] a2 E[~ x ]2 = a2 2 : The …rst line is the de…nition of variance, the second factors out the a, the third works because the expectations operator is a linear operator, and the third is the de…nition of 2 . 6. We know that E[(x x) 2 ]= 2 x We want to …nd 2 y Note that yields y =3 = E[(y y) 2 ] 1, and that y = 3x x 2 y = E[(y y) = = = = = 1 3 E[(3x E[(3x E[9(x 9E[(x 9 2x : 2 1. Substituting these in ] (3 2 x) ] 2 x) ] 2 x) ] x 1))2 ] CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 262 7. The mean is = 3 + 12 y. The variance is therefore 1 [6 2 1 2 y = 4 2 = 1 1 (3 + y)]2 + [y 2 2 1 (3 + y)]2 2 3y + 9: The derivative with respect to y is d 2 1 = y dy 2 8. All we have to do is show that G(2) (x) G(2) (x) 3: G(1) (x) for all x. We have G(1) (x) = nF n 1 (x)(1 F (x)) + F n (x) = nF n 1 (x)(1 F (x)) 0: [F n (x)] Solutions for Chapter 14 1. (a) F (x; y) x~ = 1 x~ = 2 x~ = 3 x~ = 4 y~ = 10 :04 :11 :13 :14 y~ = 20 :04 :11 :24 :37 y~ = 30 :24 :49 :69 1:00 (b) Fx~ is given by the last column of part (a), and Fy~ is given by the bottom row of part (a). (c) fx~ (1) = :24, fx~ (2) = :25, fx~ (3) = :20, fx~ (4) = :31. fy~(10) = :14, fy~(20) = :23, fy~(30) = :63: Similarly, (d) The formula for conditional density is f (xj~ y = 20) = f (x; 20)=fy~(20), which gives us f (1j~ y = 20) = 0=:23 = 0, f (2j~ y = 20) = 0, f (3j~ y = 20) = 11=23, and f (4j~ y = 20) = 12=23. (e) Using the marginal density from part (c), the mean is y = (:14)(10) + (:23)(20) + (:63)(30) = 24:9 CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 263 (f) Using part (d), xjy=20 = (0)(1) + (0)(2) + (11=23)(3) + (12=23)(4) = 81=23 (g) No. For the two to be independent we need f (x; y) = fx~ (x)fy (y). This does not hold. For example, we have f (3; 20) = :11, fx~ (3) = :20, and fy~(20) = :23, which makes fx~ (3)fy~(20) = :046 6= :11. (h) We have :04(1) + :07(2) + :02(3) + :01(4) = 2:0 :14 0(1) + 0(2) + :11(3) + :12(4) Ex [~ xj~ y = 20] = = 3:52 :23 :2(1) + :18(2) + :07(3) + :18(4) = 2:36 Ex [~ xj~ y = 30] = :63 Ey [Ex [~ xjy] = (:14)(2:0) + (:23)(3:52) + (:63)(2:36) = 2:58 Ex [~ xj~ y = 10] = Finally, using the marginal density from part (c) yields Ex [~ x] = (:24)(1) + (:25)(2) + (:20)(3) + (:31)(4) = 2:58. It works. 2. (a) f (x; y) x~ = 1 x~ = 2 x~ = 3 x~ = 4 y~ = 3 0.03 0.05 0.10 0.17 y~ = 8 0.05 0.19 0.25 0.43 y~ = 10 0.25 0.44 0.71 1.00 (b) Fx~ (1) = 0:25, Fx~ (2) = 0:44, Fx~ (3) = 0:71, Fx~ (4) = 1:00 and Fy~(3) = 0:17, Fy~(8) = 0:43, Fy~(10) = 1:00. (c) fx~ (1) = 0:25, fx~ (2) = 0:19, fx~ (3) = 0:27, fx~ (4) = 0:29 and fy~(3) = 0:17, fy~(8) = 0:26, fy~(10) = 0:57. (d) f (~ y = 3j~ x = 1) = 0:12, f (~ y = 8j~ x = 1) = 0:08, and f (~ y = 10j~ x= 1) = 0:80. (e) x = 2:6 and y = 8:29. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 264 (f) E[~ xj~ y = 3] = [(0:03)(1) + (0:02)(2) + (0:05)(3) + (0:07)(4)]=0:17 = 2: 94: (g) No. The following table shows the entries for fx~ (x)fy~(y): fx~ (x)fy~(y) x~ = 1 x~ = 2 x~ = 3 x~ = 4 y~ = 3 0.043 0.032 0.046 0.049 y~ = 8 0.065 0.049 0.070 0.075 y~ = 10 0.143 0.108 0.154 0.165 None of the entries are the same as those in the f (x; y) table. (h) 2 x = 1:32 and (i) Cov(~ x; y~) = (j) xy = 2 y = 6:45: 0:514 :176: (k) We have Ex [~ xj~ y= Ex [~ xj~ y= Ex [~ xj~ y= (:03)(1) + (:02)(2) + (:05)(3) + (:07)(4) = 2: 94 :17 (:02)(1) + (:12)(2) + (:01)(3) + (:11)(4) 8] = = 2: 81 :26 (:2)(1) + (:05)(2) + (:21)(3) + (:11)(4) 10] = = 2:40 :57 3] = Ey [Ex [~ xjy]] = (:17)(2:94) + (:26)(2:81) + (:57)(2:40) = 2:6 which is the same as the mean of x~ found above. 3. The uniform distribution over [a; b] is F (x) = x b a a when x 2 [a; b], it is 1 for x > b, and 0 for x < a. distribution is F (xjx F (x) c) = = F (c) x b c b a a a a = x c The conditional a a for x 2 [a; c], it is 1 for x > c, and 0 for x < a. But this is just the uniform distribution over [a; c]. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 265 4. x~ and y~ are independent if f (x; y) = fx~ (x)fy~(y) or, equivalently, if f (xjy) = f (x). This answer uses the latter formulation. We can see that f (~ x = 1j~ y = 10) = 1=4, and for x~ and y~ to be independent it must also be the case that f (~ x = 1j~ y = 20) = 1=4. But f (~ x= 1j~ y = 20) = a : a+b We also know that a + b must equal 0.6 so that the probabilities sum to one. Thus, a 1 a = = a+b 0:6 4 :6 a = = 0:15 4 b = 0:6 a = 0:45: 18.1 Solutions for Chapter 15 1. x = 4 and s2 = 24. Solutions for Chapter 17 1. (a) Compute the t-statistic t= x 60:02 0 p = p = 7:41 s= n 44:37= 30 which has 29 degrees of freedom. Use the Excel formula =TDIST(7.41, 29, 2) to get the p-value of 0.0000000365. The data reject the hypothesis. (b) The t-statistic is 3.706, the p-value is 0.000882, and the hypothesis is rejected. (c) The t-statistic is -0.614, the p-value for the one-tailed test is 0.27, and the hypothesis is supported. CHAPTER 18. SOLUTIONS TO END-OF-CHAPTER PROBLEMS 266 (d) The sample mean is 60.02 which is smaller than 100, so the hypothesis is supported. 2. (a) The best estimate of is the sample mean x and the best estimate of 2 is the sample variance s2 . x = 44:2 s2 = 653:8 (b) Compute the t-statistic t= 44:2 40 x p = p = 0:73 s= n 25:6= 20 and the t-statistic has 19 degrees of freedom. From here compute the p-value of 2(1 TDist(0:73; 19)) = 0:47 > 0:05 and the data support the hypothesis. (c) Compute the t-statistic t= x 44:2 60 p = p = s= n 25:6= 20 2: 76 and again the t-statistic has 19 degrees of freedom. From here compute the p-value of 2(1 TDist(2:76; 19)) = :01 25 < 0:05 and the data reject the hypothesis. INDEX alternative hypothesis, 202 astrology, 15, 29 asymptotic theory, 192 almost sure convergence, 192 Central Limit Theorem, 193 convergence in distribution, 193 convergence in probability, 192 Law of Large Numbers, 192 auctions, 161 augmented matrix, 87 Bayes’ rule, 137 likelihood, 138 posterior, 138 prior, 138 Bernoulli distribution, 145 better-than set, 126 binding constraint, 54 binomial distribution, 145 Excel command, 147 sampling from, 200 capacity constraint, 54 cdf, 145 Central Limit Theorem, 193 chain rule for functions of one variable, 13 for functions of several variables, 25 chi-square distribution, 194 relationship to normal distribution, 195 Cobb-Douglas function, 43 production function, 46 utility function, 43 cofactor, 78, 82 coin ‡ipping, 145 column space, 89, 100 comparative statics analysis, 5, 29, 31, 45, 96 267 268 INDEX implicit di¤erentiation approach, 30 total di¤erential approach, 32 complementary slackness, 57 component of a vector, 22 concave function, 120 de…nition, 121 conditional density, 177 continuous case, 177 discrete case, 177 general formula, 177 conditional expectation, 181 conditional probability, 136, 177 con…dence level, 203 constant function, 10 continuous random variable, 143 convergence of random variables, 192 almost sure, 192 in distribution, 193 in probability, 192 convex combination, 120 convex function, 122 de…nition, 122 convex set, 124 coordinate vector, 25, 81 correlation coe¢cient, 179, 180 covariance, 179 Cramer’s rule, 79, 81, 97 critical region, 202 cumulative density function, 145 degrees of freedom, 191 density function, 144 derivative, 9 chain rule, 13 division rule, 14 partial, 24 product rule, 12 determinant, 78 di¤erentiation, implicit, 29 discrete random variable, 143 distribution function, 144 dot product, 23 dynamic system, 102, 104 in matrix form, 104 e, 17 econometrics, 98, 99 eigenvalue, 105 eigenvector, 106 Euclidean space, 23 event, 132, 143 Excel commands binomial distribution, 147 t distribution, 207 expectation, 164, 165 conditional, 181 expectation operator, 165 expected value, 164 conditional, 181 experiment, 131 exponential distribution, 149 exponential function, 17 derivative, 18 F distribution, 199 relation to t distribution, 199 …rst-order condition for multidimensional optimization, 28 …rst-order condition (FOC), 15 for equality-constrained optimization, 40 for inequality-constrained optimization, 57, 65 Kuhn-Tucker conditions, 67 INDEX 269 …rst-order stochastic dominance, 159 Kuhn-Tucker Lagrangian, 66 gradient, 117 Lagrange multiplier, 39 interpretation, 43, 48, 55 Hessian, 117 Lagrangian, 39 hypothesis Kuhn-Tucker, 66 alternative, 202 lame examples, 53, 59, 99, 103 null, 202 Law of Iterated Expectations, 184 hypothesis testing, 202 Law of Large Numbers, 192 con…dence level, 203 Strong Law, 192 critical region, 202 Weak Law, 192 errors, 203 least squares, 98, 100 one-tailed, 205 projection matrix, 101 p-value, 206 Leibniz’s rule, 160, 162 signi…cance level, 203 likelihood, 138 two-tailed, 205 linear approximation, 114 linear combination, 89 identity matrix, 75 IID (independent, identically distrib- linear operator, 155 linear programming, 62 uted), 187 implicit di¤erentiation, 29, 31, 37, 97 linearly dependent vectors, 91 linearly independent vectors, 91, 92 independence, statistical, 140 logarithm, 17 independent random variables, 178 derivative, 17 independent, identically distributed, logistic distribution, 151 187 lognormal distribution, 151 inequality constraint lower animals, separation from, 156 binding, 54 nonbinding, 55 Maple commands, 207 slack, 55 marginal density function, 177 inner product, 23 martrix integration by parts, 157, 158 diagonal elements, 73 inverse matrix, 76 matrix, 72 2 2, 82 addition, 73 existence, 81 augmented, 87, 92 formula, 82 determinant, 78 IS-LM analysis, 95 dimensions, 72 Hessian, 117 joint distribution function, 175 idempotent, 102 Kuhn-Tucker conditions, 67 270 INDEX identity matrix, 75 inverse, 76 left-multiplication, 75 multiplication, 73 negative semide…nite, 118, 119 nonsingular, 81 positive semide…nite, 119 rank, 88 right-multiplication, 75 scalar multiplication, 73 singular, 81 square, 73 transpose, 75 mean, 165 sample mean, 188 standardized, 193 Monty Hall problem, 139 multivariate distribution, 175 expectation, 178 mutually exclusive events, 132 orthogonal, 101 outcome (of an experiment), 131 objective function, 30 one-tailed test, 205 order statistics, 169 …rst order statistic, 169, 170 for uniform distribution, 170, 172 second order statistic, 171, 172 IID, 187 rank (of a matrix), 88, 92 realization of a random variable, 143 rigor, 3 row-echelon decomposition, 87, 88, 92 row-echelon form, 87 p-value, 206 partial derivative, 24 cross partial, 25 second partial, 25 pdf, 145 population, 187 positive semide…nite matrix, 119 posterior probability, 138 prior probability, 138 probability density function, 145 probability distributions binomial, 145 chi-square, 194 F, 199 logistic, 151 lognormal, 151 normal, 148 t, 198 negative semide…nite matrix, 118, 119 uniform, 147 nonbinding constraint, 55 probability measure, 132, 143 nonconvex set, 124 properties, 132 norm, 24 quasiconcave, 126 normal distribution, 148 quasiconvex, 127 mean, 165 sampling from, 197 random sample, 187 standard normal, 148 random variable, 143, 164 variance, 168 continuous, 143 null hypothesis, 202 discrete, 143 271 INDEX sample, 187 sample frequency, 200 sample mean, 188 standardized, 193 variance of, 188 sample space, 131 sample variance, 189, 190 mean of, 191 scalar, 11, 23 scalar multiplication, 23 search, 181 second-order condition (SOC), 16, 116 for a function of m variables, 118, 119 second-price auction, 161, 163 signi…cance level, 203 span, 89, 92 stability conditions, 103 using eigenvalues, 109 stable process, 103 standard deviation, 167 standardized mean, 193 statistic, 187 statistical independence, 140, 178 and correlation coe¢cient, 179 and covariance, 179 statistical test, 202 submatrix, 78 support, 145 system of equations, 76, 86 Cramer’s rule, 79, 81, 97 existence and number of solutions, 91 graphing in (x; y) space, 89 graphing in column space, 89 in matrix form, 76, 97 inverse approach, 87 row-echelon decomposition approach, 88 t distribution, 198 relation to F distribution, 199 t-statistic, 199 Taylor approximation, 115 for a function of m variables, 117 test statistic, 202 told you so, 48 total di¤erential, 31 transpose, 75, 98 trial, 131 two-tailed test, 205 type I error, 203 type II error, 203 unbiased, 191 uniform distribution, 147 …rst order statistic, 170 mean, 165 second order statistic, 172 variance, 167 univariate distribution function, 175 variance, 166 vector, 22 coordinate, 25 dimension, 23 in matrix notation, 73 inequalities for ordering, 24 worse-than set, 127 Young’s Theorem, 25