The Choice of When to Buy and When To Sell

The Choice of When to Buy and When To Sell Amihai Glazer∗ Refael Hassin†‡ Irit Nowik§ arXiv:1912.02869v1 [econ.GN] 5 Dec 2019 December 9, 2019 Abstract A consumer who wants to consume a good at a particular period may nevertheless attempt to buy it earlier if he is concerned that the good will otherwise be sold. We analyze the behavior of consumers in equilibrium and the price a profit-maximizing firm would charge. We show that a firm profits by not selling early. If, however, the firm is obligated to also offer the good early, then the firm may maximize profits by setting a price which induces consumers to all arrive early, or all arrive late, depending on the good’s value to the customer. ∗ Department of Economics, University of California, Irvine. [email protected] Statistics and Operations Research, Tel Aviv University Tel Aviv 69978, Israel. [email protected] ‡ This research was supported by the Israel Science Foundation (grant No. 355/15) § Department of Industrial Engineering and Management, Jerusalem College of Technology, Jerusalem, Israel. [email protected] † 1 1 Introduction It would appear that if a firm has no cost of staying open to accept orders, it will open as early as possible. But we shall see that is not necessarily true. Suppose that all consumers prefer consuming the good in period 1. But if the store opens in period 0, and only one unit of the good is available, some consumers, fearing that the good may be unavailable in period 1, may go to the store in period 0, and buy the good early. By opening the store only in period 1, the firm precludes such wasteful behavior. The paper shows that the monopolist himself does not gain by offering the good early. Of special interest to us is rent seeking behavior—will consumers buy the good earlier than at the time they most value consuming the good because they fear a stockout. Such rent seeking behavior is common. People have been known to wait in line for hours to buy a new model of iPhone, before others do.1 They have also camped out for days and nights in the hope of buying a house2 Perhaps firms create such artificial shortages, or set prices that generate excess demand, to publicize the good, or to create a buzz about it. We do not examine such motives, but our model does examine how the costs consumers incur in seeking the good affect demand for the good and the price the firm can charge. A contribution of this paper is to shed light on inventory problems. If the firm is known to offer the good only in one period it never holds any inventory. So examining inventory issues requires asking why the firm offers the good in multiple periods. Related issues are examined by Antoniou and Fiocco (2019) who investigate the inventory behavior of a firm faced with forwardlooking consumers who can store a good in anticipation of higher future prices. They show that a seller who cannot commit to future prices can profit from holding inventory when buyers may stockpile. Other work, well surveyed by Antoniou and Fiocco (2019), documents buyer stockpiling in anticipation of higher future prices. But in that work, unlike in ours, a consumer who buys early does not affect the welfare of a consumer who intends to buy later. A further contribution of the model is to examine a seller’s decision of when to offer the good for sale, and thus to allow examination of how governmental regulations (such as blue laws, or hour restrictions on stores) affect prices or sales. 1 See https://www.cnbc.com/2019/09/20/apple-iphone-11-goes-on-sale-with-lines-outside-major-stores-around-the-wor html 2 See “House buyers sleep on street for NINE nights to buy homes.” https://www.mirror.co.uk/news/uk-news/ people-camping-out-cars-first-16230962 2 2 Literature A large literature considers dynamic pricing, in which a seller changes the price over period for the purpose of price discrimination—high-valuers buy the good early at a high price, whereas low valuers buy the good later at a low price. Under some conditions profit-maximization may require prices to decline over period, as in Su (2007), where consumers differ both in willingness to pay and in willingness to wait. Other work also considers a firm which has a fixed inventory to sell over an infinite horizon rather than over two periods (Gallien 2006). A firm may also profit by creating shortages in future periods—the possible shortage induces consumers who highly value the good to buy it at a high price, expecting a possible shortage in future periods. Analyses have considered such strategies when consumers observe the stock of inventory at the period they may buy, and when they do not. Papers considering observable stocks, and sales over two periods, include Liu and van Ryzin (2008), and Zhang and Cooper (2008). A similar model, but with customers not observing the seller’s stock, is Gallego, Phillips, and Sahin (2008). Advance sales, that is a sale made before the item is delivered, with the possibility of resale, is studied by Cachon and Feldman (2018). Their model focuses on the different prices the firm may charge in different periods, and on a consumer who learns over time his valuation of the good. Like us, they allow for the possibility that a consumer who does not buy in an early period may find that the good has been sold to someone else at an earlier period. But, in contrast to our analysis, they do not consider consumers who differ in the period at which they most value the good, and do not consider how one consumer may change his behavior depending on what other consumers do. Little work considers a firm’s strategy when it wants customers to hold inventory instead of having the firm incur the expense of holding inventory. An important, and early, paper is Blattberg, Eppen, and Lieberman (1981). They describe an inventory model in which both consumers and the retailer minimize their own costs, with the variations in price inducing customers to buy early. Glazer and Hassin (1986) consider shifting consumer demand when their holding costs are higher or lower than the firm’s. Anily and Hassin (2013) extended Glazer and Hassin’s model to heterogeneous consumers. 3 Assumptions The monopolist can sell one unit of the good, either in period 0 or in period 1. All consumers most value the good in period 1; nevertheless a consumer can buy the good in period 0 and consume 3 it then. Alternatively, we can think of a consumer buying the good in period 0, holding it until period 1, consuming then, but incurring holding costs. Such early buying reduces the consumer’s utility from V to V − K. This cost K can arise because early buying causes the consumer to incur a holding cost. Or, as with ice cream, the store can keep item cold but the consumer cannot, so that a consumer must consume the good at the period he buys it. If several consumers arrive simultaneously, the good is allocated randomly to one of them. The good’s price is P . That price is fixed over the two periods, perhaps because of menu costs, perhaps because consumers would get angry if they are charged more than they had remembered from an advertisement issued by the store, perhaps because in posting a higher price for one period than for another, some consumers would recall only the higher price, and so be unwilling to go to the store. Going to the store costs a consumer a search cost c, whether he gets the good or not. Let the equilibrium probability that a consumer chooses to arrives in period t, be qt , t = 0, 1. Let Ut be the corresponding expected utility. A consumer who does not come to the store at all (or who never arrives) obtains zero utility. In each period, a consumer may decide to arrive with certainty, to never arrive, or to arrive with positive probability less than 1. And if each consumer arrives with positive probability in any period, the aggregate probability that a consumer arrives may be 1 or less than that. Some of these possibilities, however, are irrational (for example that a consumer arrives with certainty in period 0 and also arrives with certainty in period 1). That leaves seven possible outcomes. We list them as follows, where qt is the probability a consumer arrives in period t. 4 Equilibrium behavior For the probabilities qt to constitute an equilibrium, the following conditions must be satisfied: 1. A consumer who chooses never to arrive would have non-positive expected utility were he to arrive at either period. That is, q0 = q1 = 0 =⇒ U0 , U1 ≤ 0. 2. If a consumer chooses never to arrive in period 0, but chooses to arrive in period 1 with probability less than 1, then arriving in period 1 must generate 0 utility, whereas arriving in period 0 generates non-positive utility. That is, 0 < q1 < 1 and q0 = 0 =⇒ U1 = 0 and U0 ≤ 0. 3. If, however, a consumer chooses to arrive in period 0, or in period 1, or not to arrive at all, 4 all with positive probabilities, then his expected utility arriving at any period must be zero. That is, 0 < q1 , q0 < 1 and q0 + q1 < 1 =⇒ U0 = U1 = 0. 4. If a consumer chooses to arrive in period 0 with probability less than 1, but never arrives in period 1, then his expected utility when arriving in period 0 must be zero, whereas his expected utility when arriving in period 1 must be non positive. That is, 0 < q0 < 1 and q1 = 0 =⇒ U0 = 0 and U1 ≤ 0. 5. A consumer who arrives with positive probability at either period 0 or period 1 must be indifferent between arriving at these periods. And if he always wants to arrive at some period, his expected utility must be non-negative. That is, 0 < q1 , q0 < 1 and q0 + q1 = 1 =⇒ U0 = U1 ≥ 0. 6. A consumer who chooses to arrive in period 0 must expect higher utility than arriving in period 1 or of never arriving. That is, q0 = 1 =⇒ U0 ≥ max{U1 , 0}. 7. Lastly, a consumer who chooses to arrive in period 1 must enjoy utility at least as large as when arriving in period 0 or of never arriving. That is, q1 = 1 =⇒ U1 ≥ max{U0 , 0}. Let the number of potential consumers, or the number of consumers under consideration, have a Poisson distribution with intensity λ. This assumption fits the situation with a large population of potential consumers, each person wanting the good with a small probability; the Poisson distribution is then obtained as the limit of the Binomial distribution. Although all consumers more highly value the good in period 1, if a fraction q0 > 0 nevertheless arrives in period 0, then the number of consumers arriving in period 0 has a Poisson distribution with intensity λ0 ≡ λq0 . The number of consumers arriving in period 1 has a Poisson distribution with intensity λ1 ≡ λq1 . The expected utilities are ∞  X V − K − P λj0 e−λ0 V −K −P  U0 = −c + · = −c + 1 − e−λ0 , j+1 j! λ0 (1) j=0 U1 = −c + e−λ0  V −P  1 − e−λ1 . λ1 (2) Note that the expression for U1 is multiplied by the probability e−λ0 that no consumer arrives in period 0, and hence the good is still available in period 1. By L’Hôpital’s rule, limλ→0 1−e−λ λ = 1. So as λ increases from 0 to 1 the function decreases from 1 to 1 − e−1 . 5 The firm’s expected profit is Π = P (1 − e−λ0 −λ1 ). 4.1 Characterizing the different equilibria Lemma 1. There is at most one equilibrium of each type. Proof. The claim is straightforward for equilibrium types 7, 6, and 1, where q0 and q1 are explicitly defined. The type-3 equilibrium (where a consumer may choose never to arrive) requires that U0 = 0, which uniquely defines λ0 . Substituting the resulting value in U1 = 0 uniquely defines the candidate λ1 . For the type-2 equilibrium, which assumes λ0 = q0 = 0, λ1 follows as above. Similarly, the type-4 equilibrium already assumes that q1 = 0, leading to λ1 = 0, with λ0 uniquely determined by U0 = 0. The type-5 equilibrium (where a consumer is indifferent between arriving in periods 0 and 1) requires that q0 + q1 = 1, or equivalently that λ0 + λ1 = λ. Substituting this condition into the other requirement that U0 = U1 gives V V −K = 1−λ0 eλ0 −1 λ0 1−e−(1−λ0 ) . The right-hand side is a monotonic function of λ0 . Hence, if λ0 is in [0, 1] it is uniquely defined (and corresponds to an equilibrium). We now analyse the existence of equilibrium types according to the input parameters. We normalize V − P = 1 and λ = 1. To simplify the notation, call the expected number of people arriving in period 0 x ≡ λ0 ; call the expected number of people arriving to the store in period 1 y ≡ λ1 . For each type of equilibrium we present the conditions on c and K for having an equilibrium of that type. Note that we describe the feasibility region for equilibria of types 3-5 using parametric representation of the conditions on c and K. 1. A type-1 equilibrium (where no consumer ever arrives) has λ0 = λ1 = 0, U0 ≤ 0, and U1 ≤ 0, requiring that 1 − K ≤ c and that c ≥ 1. 2. A type-2 equilibrium (where no consumer arrives in period 0, but may arrive in period 1) has λ0 = 0. The condition U1 = 0 is equivalent to c = 1−e−y y , implying that c ∈ (1 − 1/e, 1). The condition U0 ≤ 0 means that K + c ≥ 1. 3. For the type-3 equilibria (where a consumer with ideal period 1 chooses to arrive in period 0, or in period 1, or not to arrive at all, all with positive probabilities), for 0 ≤ x ≤ 1 and −x 0 ≤ y ≤ 1 − x, the condition U0 = 0 reduces to c = (1 − K) 1−ex   −y (1−x) amounts to c = e−x 1−ey ∈ e−x 1−e1−x , e−x . 6 . The condition U1 = 0 4. For the type-4 equilibrium (where no consumer arrives in period 1, but consumers may arrive in period 0), for 0 ≤ x ≤ 1, the condition U0 = 0 implies that K = 1 − c 1−ex−x , and U1 ≤ 0 with q1 = 0 implies that c ≥ e−x . 5. In the type-5 equilibrium (where a consumer is indifferent about when to arrive), λ0 + λ1 = 1. Define x ≡ λ0 . For 0 ≤ λ0 ≤ 1, the condition U0 = U1 reduces to K = 1 − And the non-negativity of U1 reduces to c ≤ e−x 1−e−(1−x) 1−x x 1−e−(1−x) 1−x ex −1 . . This condition also implies that c ≤ 1 − 1e . 6. In a type-6 equilibrium (where consumers may arrive in period 0), λ0 = 1 and λ1 = 0. The conditions U0 ≥ U1 and U0 ≥ 0 reduce to (1 − K)(1 − 1e ) ≥ 1 e, or K ≤ e−2 e−1 , and c ≤ (1 − K)(1 − 1e ). 7. In a type-7 equilibrium (where consumers may arrive in period 1), λ0 = 0 and λ1 = 1. The conditions U1 ≥ U0 and U1 ≥ 0 reduce to 1 − 1 e ≥ 1 − K, or K ≥ 1 e and c ≤ 1 − 1e . Figure 1 describes the regions corresponding to the different types of equilibria. The point at the intersection of types 7,5,3 and 2 has c = 1 − 1/e and K = 1/e. The point at the intersection of types 7, 6, 5, 4, and 3 has c = 1/e and K = (e − 2)/(e − 1). From Figure 1 we conclude that for most combinations of c and K a unique equilibrium exists. However, in the narrow region where a type-5 equilibrium exists, other equilibria also exist. In the following, denote λe ≡ (λ0 , λ1 ). Example 1 Let a consumer’s cost of going to the store be c = 0.2. Let a consumer’s penalty for buying the good earlier than at his ideal period be K = 0.4. Then the pair of arrival rates at the store λ0 = 0 and λ1 = 1 is a type-7 equilibrium (where consumers may arrive in period 1). The pair λ0 = 1 and λ1 = 0 (denoted by λe = (1, 0)) is a type-6 equilibrium (an equilibrium where consumers may arrive in period 0). And the pair λ0 = 0.6305 and λ1 = 0.3995 is a type-5 equilibrium. Note that the type-7 equilibrium is efficient. A consumer who wants to consume the good in period 1 arrives in period 1. A type-6 equilibrium suggests inefficiency. Although consumers want the good in period 1, they arrive in period 0. So if the waiting cost is positive, a consumer who arrives early may incur a cost without increasing the benefit he gets from the good. Early arrival reflects rent-seeking behavior. The same applies for a type-5 equilibrium. In terms of social welfare (the aggregated utilities), note that the social welfare function, denoted as SW , satisfies SW = (1 − e−λ0 )(V − P − K) + e−λ0 (1 − e−λ1 )(V − P ) − c(λ0 + λ1 ) = 7 1 0.9 0.8 7 0.7 2 K 0.6 5,6,7 0.5 3,5,7 1 0.4 4,5,7 0.3 0.2 3 6 4 0.1 0 0 0.2 0.4 0.6 0.8 1 1.2 c Figure 1: Types of equilibria in the (c, K) space. 8 (1 − e−λ0 )0.6 + e−λ0 (1 − e−λ1 ) − 0.2(λ0 + λ1 ). Thus social welfare for the equilibria of types 7, 6 and 5 equilibria are 0.432, 0.179 and 0.056, respectively. Put differently, if the store refused to sell in period 0, consumers would be better off. Early arrival reflects behavior of strategic complements or follow the crowd (see Hassin and Haviv 2003), where a consumer’s best response tends to follow the strategy of the others. Typically in such situations there are two extreme pure-equilibrium strategies and one mixed strategy. Example 2 Let c = 0.4 and K = 0.37. Then λe = (0, 1) is a type-7 equilibrium, λe = (0.041, 0.959) is a type 5 equilibrium, and λe = (0.989, 0) is a type-4 equilibrium. Example 3 Let c = 0.4 and K = 0.4. Then λe = (0, 1) is a type-7 equilibrium, λe = (0.63, 0.37) is a type-5 equilibrium; λe = (0.8742, 0.085) is a type-3 equilibrium. We now change the way we normalize the parameters and assume (instead of V − P = 1) that c = 1. To simplify the presentation, we use V to represent V − P . 1. In a type-1 equilibrium λ0 = λ1 = 0. The utilities U0 and U1 will both be non-positive only if V − K ≤ 1 and V ≤ 1. 2. A type-2 equilibrium has λ0 = 0. The condition U1 = 0 is equivalent to V = y , 1−e−y implying that V ≥ 1. The inequality U0 ≤ 0 means that V − K ≤ 1. 3. For the type-3 equilibrium with 0 ≤ λ0 ≤ 1 and 0 ≤ λ1 ≤ 1, U0 = 0 implies that V = K+ x 1−e−x (with x ≡ λ0 and y ≡ λ1 ). And U1 = 0 amounts to V = ex 1−ey−y . 4. For the type-4 equilibrium, with 0 ≤ λ0 ≤ 1, U0 = 0 implies that V = K + x . 1−e−x And U1 ≤ 0 (with q1 = 0) amounts to V ≤ ex . 5. For the type-5 equilibrium with 0 ≤ λ0 ≤ 1, U0 = U1 reduces to K V = 1− x 1−e−(1−x) 1−x ex −1 The non-negativity of U1 reduces to V ≥ ex 1−e1−x −(1−x) . 6. In a type-6 equilibrium, λ0 = 1 and λ1 = 0. The conditions U0 ≥ U1 and U0 ≥ 0 reduce to V ≥ e−1 e−2 K, and V ≥ K + e e−1 . 7. In a type-7 equilibrium, λ0 = 0 and λ1 = 1. The conditions U1 ≥ U0 and U1 ≥ 0 reduce to K≥ V e and V ≥ e e−1 . 9 2 1.8 1.6 7 1.4 K 1.2 1 1 3,5,7 2 5,6,7 0.8 4,5,7 0.6 3 0.4 0.2 0 6 4 0 0.5 1 1.5 2 2.5 3 3.5 V Figure 2: Types of equilibria in the (V, K) space. Figure 2 describes the regions corresponding to the different types of equilibria. The point at the intersection of types 7,5,3 and 2 has V = e/(e − 1) and K = 1/(e − 1). The point at the intersection of types 7,6,5,4 and 3 has V = e and K = e(e − 2)/(e − 1). 5 Unbounded potential demand The model with finite potential demand is too difficult to solve analytically. Moreover, we also showed that the equilibrium is not always unique. Therefore we consider the simpler situation with an infinite potential demand λ. Then the only possible equilibrium types are 1-4. The conditions for types 1 and 2 are as before. Assume first that the price P is zero. The line separating Regions 10 3 and 4 satisfies U0 = U1 = 0 and λ1 = 0. Thus from (1) we have U0 = (V − K) 1 − e−λ0 − c = 0. λ0 (3) −λ = 1, and utilizing (2), with U1 = 0, gives V e−λ0 = c. That is, Recalling that limλ→0 1−eλ
λ0 = ln Vc . Substituting this λ0 in (3) gives
c ln Vc . K=V − 1 − Vc We find it convenient to normalize all monetary values by considering c as the unit value. Thus we define v ≡ V c ,k≡ K c , p≡ P c, and π ≡ Π c. In particular, from the equation above we get k=v− ln v . 1 − v1 (4) We can verify that for v > 1, the right-hand side of (4) strictly increases from 0 to infinity. Thus, for every k > 0, Equation (4) has a unique solution u(k). Hence we define Definition 2. Let (u(k), k) be the borderline between Regions 3 and 4, then u = u(k) uniquely solves k =u− ln u . 1 − u1 The partition of the (v, k) plane is shown in Figure 3. Recall that v = V c . Thus, for example, the condition v ≤ 1, that defines Region 1, corresponds to V ≤ c, etc. Summarizing, in Region 1, (where 0 < v ≤ 1), because the search cost exceeds the value of consuming the good, consumers never arrive. In Region 2, (where 1 < v ≤ k + 1), the value of an early purchase is smaller than the search cost, so consumers arrive only in period 1. In Region 3, (where k + 1 < v ≤ u(k), and u(k) was defined in Definition 2), the arrival rate in period 0 is such that the expected utility of a consumer arriving then is zero. But the added benefit, k, of buying in period 1 rather than in period 0 is sufficiently large to compensate for the reduced probability of getting the good in period 1. Thus, consumers arrive in both periods. In Region 4, (where v ≥ u(k)), the small value of k means that a large arrival rate in period 0 discourages consumers from arriving in period 1. By setting a price p = P c, the net benefit of the purchase reduces to v − p. Thus, an increase in 11 2 1.8 1.6 1.4 k 1.2 1 1 2 3 0.8 0.6 4 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 3.5 v Figure 3: Types of equilibria in the (v, k) space - infinite potential demand. 12 p means moving to the left in the (v, k) space. The equilibrium, and therefore the profit, depend on the region where we end up. Summarizing the results on the four regions: • In Region 1, λ0 = λ1 = 0, U0 , U1 ≤ 0, and it is reached iff 0 ≤ v − p < 1. • In Region 2, λ0 = 0, 0 < λ1 < λ, U1 = 0, U0 ≤ 0, and it is reached iff 1 ≤ v − p ≤ k + 1. • In Region 3, λ0 , λ1 > 0, λ0 + λ1 < λ, U0 = U1 = 0, and it is reached iff k + 1 < v − p < u(k), where u was defined in Definition 2. • In Region 4, 0 < λ0 < λ, λ1 = 0, U0 = 0, U1 ≤ 0, and it is reached iff u(k) ≤ v − p ≤ v, where u(k) satisfies k = u − ln u 1 1− u (see Definition 2). Two equations are central for the model. First, the equation U0 = 0, which is satisfied in Regions 3,4, and by (3) gives v−p−k = λ0 . 1 − e−λ0 (5) Second, the equation U1 = 0 which is satisfied in Regions 2,3, and by (2) gives v−p= 5.1 λ1 e λ 0 . 1 − e−λ1 (6) Arrival rates We consider the behavior of λ0 , λ1 and λ0 + λ1 in each of the regions. Proposition 3. Let v be the value of consuming the good at the ideal period, and p the price of the good. The arrival rates (namely λ0 and λ1 ) in period 0 and 1 satisfy: • In Region 1, λ0 = λ1 = 0. • In Region 2, λ0 = 0, and λ1 increases with v − p. • In Region 3, λ0 , λ1 > 0, λ0 increases with v − p, and λ1 declines with v − p. • In Region 4, λ0 increases with v − p, and λ1 = 0. As expected, the arrivals rates λ0 and λ1 usually decline with the price. Nevertheless, an exception appears in Region 3, where λ1 declines with v − p. An explanation is that a higher price induces fewer consumers to arrive in period 0. This makes it more likely that the good is still 13 0.5 a -1.0 -0.5 0.5 1.0 -0.5 -1.0 Figure 4: The graph of W [a] for − 1e ≤ a ≤ 1. available in period 1, and so more consumers arrive in period 1, even though the price is higher. The proof of Proposition 3 appears in the Appendix (see Section 8.1). Moreover, the following theorem claims that in Region 3 (where consumers may arrive in both periods and expected consumer utility is zero) the arrival rate in period 1 increases with the price, and this increased arrival rate exceeds the reduced arrival rate in period 0. Theorem 4. Let a consumer value consuming the good more in period 1 than in period 0 (that is, k > 0). Let the value of consuming the good at the ideal period be v, and the price of the good p. Then in Region 3 (where consumers may arrive in both periods and expected consumer utility is zero) the aggregate arrival rate λ0 + λ1 decreases in v − p. Our central results involve a function called the Lambert function W [a], (also called the omega function or product logarithm). The Lambert function is actually a set of functions, namely the branches of the inverse relation of the function zez . In other words, if a = zez , then W [a] = z. Because the function zez is not injective, W [a] is multivalued (except at 0). Note that the relation is defined only for a ≥ −e−1 , and is double-valued on (−1/e, 0). The additional constraint that W [a] ≥ −1 defines a single-valued function, which is presented in Figure 4). Definition 5. For all a ≥ −e−1 : W [·] is the inverse function of aea , that satisfies W [a] ≥ −1. 14 The following properties of W will be used below. For proofs see Corless et al. (1996). • W1. W [·] is an increasing function for all a ≥ −e−1 . • W2. W [−e−1 ] = −1. • W3. W ′ [a] = W [a] a(W [a]+1) . • W4. W [0] = 0. • W5. e−W [a] = W [a] a . • W6. W [aea ] = a for all a ≥ −1. To prove Theorem 4, we first need to establish several results. Definition 6. For all a, R(a) = W [−ae−a ]. Note that by W6, for all a ≤ 1, (which is equivalent to −a ≥ −1), R(a) = −a. In Regions 2-4, however, where v − p > 1, we cannot exploit this attractive property. Proposition 7. Let v be the value of consuming the good at the ideal period 1, p the price of the good, and k the reduced utility if a person with ideal period 1 buys the good in period 0. Then in Regions 3 and 4, the arrival rate λ0 = v − p − k + R(v − p − k) uniquely solves (5). The proof of Proposition 7 appears in the Appendix (see Section 8.2.1). The proof of the following Corollary is very similar to the proof of Proposition 7. Corollary 8. Let v be the value of consuming the good at the ideal period 1, p be the price of the good, k be the reduced utility if a person with ideal period 1 buys the good in period 0, and R(v − p) = W [−(v − p)e−(v−p) ]. In Region 2 (where 1 < v ≤ k + 1), the unique arrival rate that solves (6) is λ1 = v − p + R(v − p). . Definition 9. For all a, A(a) = −(a + k)e−a+ Denote A = A(v − p − k). 15 kR(a) a . Proposition 10. Let v be the value of consuming the good at the ideal period 1, p be the price of the good, and k be the reduced utility if a person with ideal period 1 buys the good in period 0. Then in Region 3 (where consumers may arrive in both periods and expected consumer utility is zero) the unique solution to the equation system of (5) and (6) is λ0 = v − p − k + R(v − p − k), λ1 = W [A] − (v − p)R(v − p − k) . v−p−k The proof of Proposition 10, appears in the Appendix (see Sections 8.2.2). The presentation and proof of several lemmas needed for the proof of Theorem 4, appear in Section 8.2.3, and finally the proof of Theorem 4 appears in Section 8.2.4. 5.2 Profit maximization Consider the profit-maximizing price. The seller’s expected profit (in units of c) is the price p > 0, multiplied by the probability that the good is sold (which is the same as the probability of at least one arrival), thus   π = p 1 − e−(λ0 +λ1 ) . (7) For any given pair (v, k), we need to find the price that maximizes the seller’s profit. As explained, raising the price is equivalent to moving to the left of v. The first step, is to find the expression for π in each region and then the local maximum of each of these expressions. However, these local maxima may be obtained outside the region in which the corresponding expression of π is valid, and in that case it is irrelevant. In other words, the prices that are relevant are the local maximum points that satisfy that (v − p, k) still belongs to the original region of (v, k). In case of a region that has no relevant local maximum, we need to check its end points. Among these candidates of p, the p∗ that attains the maximal value for π(p), is the price that maximizes the seller’s profit. For any given pair (v, k) we now express the four regions in terms of p. At the beginning of Section 5, we defined the regions in terms of v − p. The definition of the regions in terms of p, are: • Region 1: v − 1 < p ≤ v. • Region 2: v − k − 1 ≤ p < v − 1. 16 7 6 5 4 3 4 2 3 2 1 1 2 4 6 8 10 Figure 5: π as a function of p. • Region 3: v − u < p < v − k − 1. • Region 4: 0 ≤ p ≤ v − u, where u was defined as the unique solution to k = u − ln u 1 1− u (see Definition 2). Note that not all four regions exist for all pairs of (v, k). For example, if v < 1, then (because p ≥ 0), all regions except for Region 1, are empty, and if v − k − 1 < 0, then Regions 3 and 4 are empty. See Figure 5, that presents π for v = 10 and k = 2. The numbered regions in Figure 5 correspond to the regions in Figure 3. In this example, Regions 2 and 4 each has a relevant local maximum. But in Region 3, the expression for π is monotonically increasing in p, and thus the right end of Region 3 is also a candidate. As seen, in this example the local maximum in Region 2 is the price that maximizes the seller’s profit. The monotonicity of π seen in Region 3 in this example also holds in general for all (v, k), as stated in the following proposition. Proposition 11. Let v be the value of consuming the good at the ideal period, and k the reduced utility of consuming the good at an earlier period. For given values of p and k, the firm’s profit π increases in p for all p, s.t., (v − p, k) belongs to Region 3 (where consumers may arrive in both periods and expected consumer utility is zero). 17 
Proof. By (7), π = p 1 − e−(λ0 +λ1 ) . Now, by Theorem 4, in Region 3, λ0 + λ1 declines in v − p,
hence λ0 + λ1 increases in p. This implies that π = p 1 − e−(λ0 +λ1 ) also increases in p. It follows from Proposition 11 that in general, the global maximum can never be attained inside Region 3. However, it may be attained at its right end v − k − 1. As we will show, in general, Regions 2 and 4 do not always have relevant local maxima. Denote the right end of Region 3 as p3 Then p3 = v − k − 1. Denote p2 and p4 as the local maxima of Regions 2 and 4 respectively. Recall that if p2 does not belong to Region 2, then it is irrelevant. Similarly, for p4 and Region 4. In contrast, p3 , which is the point separating between Regions 2 and 3, always exists and is relevant. Definition 12. W = W [−(k + 1)e−(k+1) ]. Denote π(pi ) = πi , i = 1, 2, 3. Proposition 13. Let v be the value of consuming the good at the ideal period 1, p the price of the good, and k the reduced utility if a person with ideal period 1 buys the good in period 0. 1. p2 = v − ln (v) , 1 − v1 π2 = v − 1 − ln (v) . (8) π4 = v − k − 1 − ln (v − k) , (9) 2. p4 = v − k − ln (v − k) 1 , 1 − v−k 3. p3 = v − k − 1,   π3 = (v − k − 1) 1 − e−(W +k+1) . (10) The proof of Proposition 13 appears in the Appendix (see Section 8.3.1). Theorem 14. For all p ≥ 0: 1. π4 < π2 2. π3 ≤ π2 , where πi = π(pi ), pi is the local maximum of the expression of π in Region i = 1, 2, and π3 = π(p3 ), where p3 = v − k − 1 is the border point between Regions 2 and 3. 18 The proof of Theorem 14 appears in the Appendix (see Section 8.3.2). The following Corollary follows from Theorem 14. Corollary 15. If p2 is in Region 2, then the price that maximizes π is p2 . We now find the conditions that guarantee that p2 is in Region 2. Lemma 16. If 1 ≤ v ≤ eW +k+1 , then p2 belongs to Region 2. The proof of Lemma 16 appears in the Appendix (see Section 8.3.3). Corollary 17. If 1 ≤ v ≤ eW +k+1 , then p∗ = p2 = v − ln(v) , 1− v1 and π ∗ = π2 = v − 1 − ln (v) . For larger v, namely v ≥ eW +k+1 , we need to find where π3 ≥ π4 is satisfied. In that case, p∗ = p3 . Recall that p3 is the point that separates between Regions 2 and 3, therefore it always exists and is relevant. But p4 , which is the local maximum of the expression for π in Region 4, may not belong to Region 4, and in that case it is not relevant.
Recall that π4 = v − k − 1 − ln (v − k), and π3 = (v − k − 1) 1 − e−(W +k+1) ). Let f (v) = π4 − π3 . Hence   v W − ln (v − k). Definition 18. f (v) = 1 − k+1 Denote vm as the minimum of f in Region 3. Lemma 19. Given k, f (v) is a strictly convex function which has exactly two roots, the smaller of which is k + 1. The proof of Lemma 19 appears in the Appendix (see Section 8.3.4). Definition 20. vf is the unique value that satisfies both: f (v) = 0, and v > k + 1. See Figure 6 that presents f as a function of v, for k = 1. Since f (v) = π4 − π3 , it follows that for all v satisfying k + 1 < v < vf , π4 < π3 . Recall that by Theorem 14, π3 ≤ π2 , and that by Lemma 16, p2 is relevant for all v satisfying 1 ≤ v ≤ eW +k+1 . Corollary 21. Let k be the cost of an early arrival, let W stand for W [−(k + 1)e−(k+1) ], and let vf be the larger root of f = π4 − π3 . Then for every pair (v, k) eW +k+1 ≤ vf . The proof of Corollary 21 appears in the Appendix (see Section 8.3.5). 19 f 2.0 1.5 1.0 0.5 k+1 vf 5 10 15 20 v -0.5 Figure 6: f as a function of v Corollary 22. • For all 1 ≤ v ≤ eW +k+1 , • For all eW +k+1 < v ≤ vf , p∗ = p2 , and π ∗ = π2 . p∗ = p3 , and π ∗ = π3 . But what happens for v > vf ? By Lemma 19, π4 > π3 for v > vf . But, is p4 relevant (i.e., belongs to Region 4), for v > vf ? The following Lemma says Yes. Lemma 23. 1. If v ≥ k + u, then p4 belongs to Region 4. 2. For every pair (v, k), vf ≥ k + u. The proof of Lemma 23 appears in the Appendix (see Section 8.3.6). Recall that W stands for W [−(k + 1)e−(k+1) ], and that vf is the root of f (v) that is greater than the root k + 1. The following theorem summarizes the results. Theorem 24. Let v be the value of consuming the good at the ideal period, and k the reduced utility of consuming the good at an earlier period. Then for given values of v and k, the profit-maximizing (normalized) price (p∗ ), the induced arrival rates (λ0 and λ1 ), and the associated expected (normalized) profit (π ∗ ) are 20 1. If v < 1, then λ0 = 0, λ1 = 0, and π ∗ = 0. 2. If 1 ≤ v ≤ eW +k+1 , then p∗ = v − ln v 1− v1 , π ∗ = v − 1 − ln v, λ0 = 0, and λ1 = ln v.
3. If eW +k+1 < v ≤ vf , then p∗ = v − k − 1, π ∗ = (v − k − 1) 1 − e−(W +k+1) , λ0 = 0, and λ1 = W + k + 1. 4. If v > vf , then p∗ = v − k − ln(v−k) 1 , 1− v−k π ∗ = v − k − 1 − ln (v − k), λ0 = ln (v − k), and λ1 = 0, where W = W [−(k + 1)e−(k+1) ]. The profit-maximizing strategy may have the firm set the price such that consumers buy only in period 1 (cases 2 and 3) or only in period 0 (case 4). The firm’s profit depends on the value of k (the penalty for hiring early). The firm’s profits are not maximized when k = 0, but rather when k makes the value of eW +k+1 greater than v. Profits are then π2 = v − 1 − ln v, which, according to Theorem 14, is the maximum profit. Note that this conclusion relates to the case in which the firm must offer the good at both periods. But as the following section shows, the firm is always better off not offering the good early. 5.3 Outcomes with a single period Consider a model with a single-period, in which consumers are only allowed to come in period 1. This is equivalent to making k infinite in the original model with two periods. Then, unless v < 1 (which results in π = 0), limk→∞ eW +k+1 = ∞. And so 1 ≤ v < eW +k+1 , thus belonging to the second case in Theorem 24. Hence π ∗ = π2 = v − 1 − ln v. By Theorem 14, π2 ≥ π3 , π4 . Hence the seller’s highest profit is when selling the good in period 1. It does not offer the good at an earlier period than the period most desired by the consumers. 5.4 Social optimization Clearly, social welfare is optimized when consumers only buy in Period 1. The optimal arrival rate λ1 is determined by solving maxλ1 {v1 (1 − e−λ1 ) − λ1 }, which gives the optimal value λ∗1 = ln v − p. We now ask whether when the arrival rates are determined by consumers’ equilibrium behavior, is it socially optimal to forbid early sales. Denote the normalized (i.e., divided by the search cost c) social welfare functions in the singleand two-period cases as SW1 and SW2 respectively. If v < 1, then there are no arrivals, hence SW1 = SW2 = 0. Assume v ≥ 1. For the single-period case, because all consumers need to pay the 21 (normalized) search cost of 1 but at most one consumer gains the good (if indeed there is at least one arrival), then SW1 = (v − p)(1 − e−λ1 ) − λ1 . (11) As k → ∞, Regions 3 and 4 become empty, so we are left with Region 2. In Region 2, equation (6) gives v − p = λ1 . 1−e−λ1 Substituting this in (11) gives 0. For the two-period case, one may assume that larger k reduces social welfare. However, we show now that in equilibrium, social welfare is 0 for all k The reason is that in all 1-4 types of equilibrium, whenever there is a positive arrival rate, λi > 0 then the corresponding utility satisfies Ui = 0. Consider SW2 = (1 − e−λ0 )(v − p − k) + e−λ0 (1 − e−λ1 )(v − p) − (λ0 + λ1 ). (12) Proposition 25. In all equilibria, social welfare in the two-period case equals zero, namely, SW2 = 0. Proof. • In Region 1, λ0 = λ1 = 0. Substituting this in (12) gives SW2 = 0. • In Region 2, (6) with λ0 = 0, gives v − p = λ1 . 1−e−λ1 Substituting this in (12) gives SW2 = 0. • In Region 3, by (5) and (6) we have v−p−k = λ0 λ1 e λ 0 , and v − p = . 1 − e−λ0 1 − e−λ1 Substituting this in (12) again gives SW2 = 0. • In Region 4, (5) with λ1 = 0, gives v − p − k = λ0 . 1−e−λ0 Substituting this in (12) gives SW2 = 0. 6 Back to the finite case When the arrival rate λ is finite, equilibrium types 5-7 do exist. The sum of arrival rates λ0 +λ1 = λ     is fixed. This implies that the expected profit π = p 1 − e−(λ0 +λ1 ) = p 1 − e−λ is linear and 22 increasing in p. Hence the candidates for the maximum points are the border-points between the regions. For the other (former) equilibria types 1-4, we proved that for equilibrium type 3, λ0 + λ1 increases with p, so the points on the left borders of the region in Figure 2 are candidates for maximal profit. In equilibrium types 2 and 4, we have proved that there is a local maximum for π that are also candidates for maximal profit. These results still hold for the finite case. 7 Conclusion A consumer who wants to consume a good at a particular period may nevertheless attempt to buy it earlier if he is concerned that in delaying he would find the good already sold. This paper considers a model in which the good may be offered at two periods; the period in which all consumers most value the good (period 1), and an earlier period (period 0). We show that a firm profits by not selling early. The strategy of consumers, when deciding if and when to arrive, is more complicated than one may suppose, and can generate some unexpected behavior. For example, the arrival rate at period 1 may decline with the surplus a person gets from buying the good; indeed even the aggregate arrival rate (from both periods) can decline with that surplus. The behavior of the firm can also be surprising. If the firm is obligated to also offer the good early, but must charge the same price in both periods, then, depending on the value of the good to a consumer, the firm may maximize profits by setting a price which induces consumers to arrive only in period 1, or under different conditions, induces them to arrive only in period 0. The firm would not set a price which induces consumers to arrive in both periods. This result also means that, even with an infinitesimal cost of opening the store in some period, the firm will not want to remain open in all periods. In any case, under the assumptions of the model, the firm is always better off not offering the good early. 23 8 Appendix Let v1 = v − p, and let v0 = v − p − k. Then v1 is the consumer’s benefit when buying in period 1 at price p, and v0 is the consumer’s benefit when buying in period 0 at price p. Note that given v and k, the values v0 and v1 are functions of p. We find that most of the expressions appearing in our analysis, are functions of v − p − k, thus we center our attention on v0 . Summarizing the results on the four regions in terms of v0 , • In Region 1, λ0 = λ1 = 0, U0 , U1 ≤ 0, and it is reached iff −k ≤ v0 < 1 − k. • In Region 2, λ0 = 0, 0 < λ1 < λ, U1 = 0, U0 ≤ 0, and it is reached iff 1 − k ≤ v0 ≤ 1. • In Region 3, λ0 , λ1 > 0, λ0 + λ1 < λ, U0 = U1 = 0, and it is reached iff 1 < v0 < u − k. • In Region 4, 0 < λ0 < λ, λ1 = 0, U0 = 0, U1 ≤ 0, and it is reached iff u − k ≤ v0 ≤ v − k. 8.1 Proof of Proposition 3 Recall that x ≡ λ0 , y ≡ λ1 . 1. The proof of the first statement of the proposition follows immediately from the definition of Region 1. 2. In Region 2, λ0 = 0. Thus (6) becomes v1 = y . 1 − e−y Hence dv1 1 − e−y − ye−y = . dy (1 − e−y )2 Because y +1 ≤ ey , for y ≥ 0 The numerator on the right-hand side satisfies 1−e−y (1+y) ≥ 0. Thus v1′ ≥ 0 and v1 increases with y. Hence y increases with v1 (and thus in v0 ), proving the second claim of the proposition. 3. In Region 3, λ0 and λ1 are both positive, and both equations (5) and (6) are satisfied. Similarly to the proof above for y in Region 2, it follows from (5) that in Region 3, x increases with v0 . From (5) and (6) together, we have k+ x yex = . 1 − e−x 1 − e−y 24 So ke−x + x y = . ex − 1 1 − e−y (13) The left-hand side of (13) declines with x ≡ λ0 since  ke−x + x ′ ex − 1 − xex ex (x − 1) + 1 −x −x = −ke + = −ke − . ex − 1 (ex − 1)2 (ex − 1)2 The expression ex (x − 1) + 1, appearing in the numerator on the right-hand side, increases x with x and equals 0 when x = 0. Hence it is positive and so −ke−x − e (e(x−1)+1 < 0, implying x −1)2 that the left-hand side of (13) indeed decreases in x. In contrast, the right-hand side of (13) increases in y ≡ λ1 . Thus y decreases in x. Because x ≡ λ0 increases in v0 , it follows that y ≡ λ1 declines with v0 , proving the third claim of the proposition. 4. In Region 4 y ≡ λ1 = 0 and (5) is satisfied. And so, as in Region 3, x ≡ λ0 increases with v0 , proving the last statement of the proposition. 8.2 Proof of Theorem 4 To prove Theorem 4, we first need to establish several results, using the Lambert function W [x], (see Definition 5 in Section 5.1). We have also presented a list of W1-W6 properties of the Lambert function, that we use in our analysis ahead (see Section 5.1). Recall that we have denoted R(a) = W [−ae−a ]. 8.2.1 Proof of Proposition 7 First, we show that λ0 = v0 + R(v0 ) indeed solves (5). Substituting λ0 = v0 + R(v0 ) in the right-hand side of (5), gives v0 + R(v0 ) . 1 − e−v0 e−R(v0 ) (14) By W5, e−R(v0 ) = R(v0 ) . −v0 e−v0 25 (15) Substituting this in (14) gives v0 + R(v0 ) R(v0 ) v0 1+ = v0 , which is the left-hand side of (5). Now, because the right-hand side of (5) is monotonic in λ0 , then for any given v0 the value λ0 = v0 + R(v0 ) uniquely solves (5). 8.2.2 Proof of Proposition 10 Recall that A(a) = −(a + k)e−a+ kR(a) a , and that A = A(v0 ). From Proposition 7, λ0 = v0 + R(v0 ) uniquely solves (5). Substituting this in (6) gives λ1 ev0 +R(v0 ) , 1 − e−λ1 v1 = which is the same as v1 e−v0 −R(v0 ) = λ1 . 1 − e−λ1 (16) Since the right-hand side of (16) is monotonic in λ1 , for any given pair v0 , v1 at most one value of λ1 satisfies (16). We now show that the proposed solution λ1 = W [A] − Substituting this λ1 in the right-hand side of (16) gives W [A] − 1−e v1 R(v0 ) v0 v1 R(v0 ) v0 . e−W [A] Because of W5, this expression is equal to W [A] − 1− e v1 R(v0 ) v0 v1 R(v0 ) v0 W [A] , A which is equal to  W [A] − A−e v1 R(v0 ) v0 v1 R(v0 ) v0  A . W [A] Since v1 = v0 + k, the above equals  v1 R(v0 ) v0 −v1 e  kR(v0 ) −v + − W [A] v1 e 0 v0 −v0 + kR(v0 ) v0 −e 26 v1 R(v0 ) v0 W [A] v1 R(v0 ) v0 satisfies (16). =   kR(v ) R(v )+ v 0 0 − W [A] e 0 v1 R(v0 ) v0 −v1 e −v0 + kR(v0 ) v0 −e v1 R(v0 ) v0 · v1 e−v0 −R(v0 ) W [A] Hence we need to prove that the quotient above equals 1, namely that   v R(v ) v1 R(v0 ) kR(v0 ) kR(v ) R(v )+ v 0 −v + 1 0 0 = −v1 e 0 v0 − e v0 W [A]. − W [A] e 0 v0 Note that the expression e R(v0 )+ e kR(v0 ) v0 R(v0 )+ (17) appearing on the left side satisfies kR(v0 ) v0 =e v0 +k v0
R(v0 ) =e v1 R(v0 ) v0 . Substituting this in the left-hand side of (17) gives v1 R(v0 ) 0) v1 R(v0 ) v1 R(v e v0 − e v0 W [A]. v0 Hence we only need to prove that kR(v0 ) 0) v1 R(v0 ) v1 R(v −v + e v0 = −v1 e 0 v0 . v0 This is equivalent to proving that 0 ) +v − kR(v0 ) R(v0 ) v1 R(v 0 v0 e v0 = −1. v0 Note that the left-hand side of the above equation equals 0 ) +v R(v0 ) (v1 −k)R(v R(v0 ) R(v0 )+v0 0 v0 e = e . v0 v0 By (15) the above equals R(v0 ) (−v0 e−v0 )ev0 = −1. v0 R(v0 ) 8.2.3 Several lemmas for the proof of Theorem 4 Recall Definition 2, that u = u(k) satisfies k = u − ln u 1 1− u . Note that u − k = increasing in u, which is strictly increasing in k. The following lemma proves essential properties of the function R(v0 ). 27 ln u 1 1− u , is strictly Lemma 26. • R1. R(v0 ) is negative and increasing for all v0 > 1. • R2. R(1) = −1. • R3. R(u − k) = • R4. R′ (v0 ) = k u − 1. −R(v0 ) R(v0 )+1 · v0 −1 v0 . Proof. 1. Since −v0 < −1 < 0, then −v0 e−v0 < 0, and by W1 and W4, R(v0 ) = W [−v0 e−v0 ] < 0. Now, (−v0 e−v0 )′ = e−v0 (v0 − 1) > 0 (since 1 < v0 ). The Lambert function is increasing and so W [−v0 e−v0 ] increases with v0 , proving R1. 2. To prove R2, note that by W2 R(1) = W [−e−1 ] = −1. 3. We wish to prove that R(u − k) = k u − 1. Recall that u satisfies k = u − ln u 1 1− u . Hence (k − u)(1 − u1 ) = − ln u, and so k − u = − ln u + ( uk − 1). Adding ln (k − u) to both sides of the equation gives k  k  −1 + −1 . u u (k − u) + ln (k − u) = ln Thus (k − u)ek−u = k u  k − 1 e u −1 . Applying the Lambert function to both sides of the equation gives W [(k − u)ek−u ] = W h k u  k i − 1 e u −1 . The left-hand side of (18) is R(u − k). Now by W6, since W k u (18) − 1 ≥ −1 then h k  k i k − 1 e u −1 = − 1, u u proving R3. 4. R′ (v0 ) = W ′ [−v0 e−v0 ]e−v0 (v0 − 1). 28 (19) Hence by W3, R′ (v0 ) = −R(v0 ) v0 − 1 R(v0 )e−v0 (v0 − 1) = · , −v0 e−v0 (R(v0 ) + 1) R(v0 ) + 1 v0 proving R4. For the next result we need the following lemma. Lemma 27. The expression v0 + (v0 + k)R(v0 ), is negative in Region 3. Proof. Substituting v0 = u−k, which is the right end of Region 3, in: v0 +(v0 +k)R(v0 ), and recalling that by R3, R(u−k) = k u −1, gives 0. Additionally, by R4, the derivative 1+R(v0 )+(v0 +k)R′ (v0 ), of this expression is 1 + R(v0 ) + −R(v0 )(v0 + k)(v0 − 1) . v0 (R(v0 ) + 1) By R1 and R2, in Region 3: 1 + R(v0 ) > 0, and −R(v0 ) > 0. Hence the derivative is positive and so the expression v0 + (v0 + k)R(v0 ), is negative in Region 3. Recall Definition 9, that A = A(v0 ) = −(v0 + k)e −v0 + kR(v0 ) v0 , and recall that u = u(k). Lemma 28. 1. W [A(u − k)] = −1. 2. A(v0 ) is decreasing for 1 < v0 < u − k. Proof. 1. Since W is strictly monotonic and by W2, W [−e−1 ] = −1, we must prove that A(u − k) = −e−1 . Now, A(u − k) = −uek−u+ From R3, R(u − k) = k u kR(u−k) u−k . (20) − 1. Substituting this in the right-hand side of (20) gives −uek−u+ k(k/u−1) u−k k k = −uek−u− u = −ek−u− u +ln u ,   Substituting ln u = −(k − u) 1 − u1 gives −e  k 1 k−u− u −(k−u) 1− u 29  = −e−1 . 2. To prove that A(v0 ) is decreasing, we arrive at A′ (v0 ) = e −v0 + kR v 0  v0 (v0 + k − 1) + kR v02 (R + 1)  v0 + (v0 + k)R
, (21) where R = R(v0 ). By R1 and R2, the denominator is positive. Thus we need to prove that the numerator is negative. The expression in the first parentheses is positive since v0 (v0 + k − 1) + kR = k(v0 + R) + v0 (v0 − 1) > 0, since in Region 3, v0 > 1, and v0 + R > 1 + R, (which is positive according to R1 and R2). The expression appearing in the second parentheses is negative by Lemma 27, thus A′ (v0 ) is negative implying that A(v0 ) is indeed decreasing there. Recall that R = R(v0 ), and that u = u(k) is defined by the equation k = u − ln u 1 1− u . We wish to look at W [A] as a function of k. Given v0 , denote by k0 , the k that satisfies v0 = u(k0 ) − k0 , and denote u0 = u(k0 ). Also denote by A0 the A that correspond to k0 , namely A0 = −(v0 +k0 )e −v0 + k0 R v0 . Lemma 29. Given v0 : 1. W [A] is increasing in k, in Region 3. 2. d dk W [A] 3. d dk W [A] k=k 0 is decreasing in k, in Region 3. = 1 u0 . Proof. 1. To prove that W [A] is increasing in k in Region 3, note that By Lemma 27, v0 + (v0 + k)R < 0 in Region 3, thus dA dk dA dk = −e −v0 + kR v0 v0 (v0 + (v0 + k)R). > 0 and A is increasing in k in Region 3. Since W is an increasing function, then W [A] is also increasing in k in Region 3. 2. Using W3, we arrive at d (v0 + (v0 + k)R)W [A] W [A] = . dk v0 (v0 + k)(W [A] + 1) The above equals v0 + (v0 + k)R (v0 + (v0 + k)R) − . v0 (v0 + k) v0 (v0 + k)(W [A] + 1) 30 (22) Differentiating the above expression with respect to k gives d v0 (v0 + k) dk W [A](v0 + (v0 + k)R) −v02 d2 W [A] = + . dk 2 (v0 (v0 + k))2 (v0 (v0 + k)(W [A] + 1))2 (23) The first term in (23) is negative. The second term is also negative since we proved in Part 1, that d dk W [A] is positive, and by Lemma 27, v0 + (v0 + k)R is negative. 3. By (22) (v0 + (v0 + k)R)W [A] d W [A] = . dk v0 (v0 + k)(W [A] + 1) Note that for k = k0 , the numerator vanishes according to Lemma 27, and the denominator vanishes according to Part 1 of Lemma 28. By L’Hôpital’s rule d W [A] dk  (v0 + (v0 + k)R)W [A]   = d v (v + k)(W [A] + 1) 0 0 dk d dk k=k0  k=k0 . k=k0 Hence d W [A] dk Denote s = d dk W [A] k=k , 0 k=k0 = d W [A] RW [A0 ] + (v0 + (v0 + k0 )R) dk v0 (W [A0 ] + 1) + v0 (v0 + k=k0 d k0 ) dk W [A] k=k0 . (24) then (24) becomes s= RW [A0 ] + (v0 + (v0 + k0 )R)s . v0 (W [A0 ] + 1) + v0 (v0 + k0 )s Recalling that v0 + (v0 + k0 )R appearing in the numerator equals 0, and that W [A0 ] = −1, we obtain s= −R v0 (v0 + k0 )s (25) Note that s = 0 does not solve (25), hence the above expression is well defined. From (25) we get s2 = By R3, R = R(v0 ) = R(u0 − k0 ) = k0 u0 −R . v0 (v0 + k0 ) (26) − 1. Substituting this, and also v0 = u0 − k0 , in (26) 31 gives 2 s = 1− k0 u0 (u0 − k0 )u0 = 1 u0 − k0 = 2. 2 (u0 − k0 )u0 u0 (27) Now, from the proof of the first part of Lemma 29, (that W [A] is increasing in k, in Region 3), we get that d dk W [A] is non-negative in Region 3. In particular, s ≥ 0. Combining this with the fact that s 6= 0, which we established earlier, gives s > 0. This, together with (27), implies that s = 1 u0 . We can now prove Theorem 4. 8.2.4 Proof of Theorem 4 Recall that u = u(k) is the solution for k = u − ln u 1 1− u (see Definition (2)). In Region 3, by Proposition 10, λ0 + λ1 = v0 + R(v0 ) + W [A] − v1 R(v0 ) . v0 We utilize R4 and (21), to arrive at   v02 W [A](R + 1) + (v0 + k) v02 + v0 (v0 + k)R + kR2   , (λ0 + λ1 )′ = v02 (v0 + k) W [A] + 1 (R + 1) (28) where the derivative is according to v0 . According to R1 and R2, R + 1 > 0 in Region 3 and according to Lemma 28, W [A] + 1 > 0 in Region 3. Hence, the denominator is positive in Region 3. Thus it is sufficient to prove that the numerator is negative. By substituting the end points of Region 3, namely, v0 = 1, v0 = u − k, it is easily verified that at the end points of Region 3, the numerator vanishes. We wish to prove that for all k ≥ 0, the numerator is negative for all 1 < v0 < u − k (i.e., for all v0 in Region 3).This is illustrated in Figure 7 which presents the numerator as a function of v, for k = 2. Note that u(2) = 3.81449, hence Region 3 is 1 < v0 < 1.81449. 32 1.2 1.4 1.6 1.8 -0.5 -1.0 -1.5 Figure 7: The numerator of (λ0 + λ1 )′ as a function of v0 , in Region 3, for k = 2. Recall that we denoted v0 = u0 − k0 . For any given v0 , if we look at the numerator as a function of k, then, as explained, for k0 = u0 − v0 (corresponding to the right end point v0 = u0 − k0 of Region 3 when k = k0 ), the numerator equals 0. We wish to prove that given v0 , for all k > k0 , the numerator is decreasing as a function of k, and therefore is negative. This will be proved shortly and is demonstrated in Figure 8. In Figure 8 which presents the numerator for k = 2, 4, 6, 8, 10, we see that at v0 = 1.81449 the numerator equals zero for k = k0 = 2, (upper line), and then for increasing k, the value of the numerator decreases (and is thus negative as claimed). Note also that the length of Region 3, increases with k. This is so, since for all k > 0, the left end of Region 3, is 1, and the right end is u − k which equals ln u 1 1− u , which increases with u and thus with k. In particular when k → 0, the right end u(k) − k → 1, and so when k = 0, Region 3 is empty (since it consists of 1 < v0 < 1). Given v0 , denote by N (k), the numerator of (λ0 + λ1 )′ (when the derivative is with respect to v0 ), namely   N (k) = v02 W [A](R + 1) + (v0 + k) v02 + v0 (v0 + k)R + kR2 . We now differentiate N (k) with respect to k. Note that v0 and k0 are fixed in k, and consequently R, which is a function of v0 only, is fixed as well. d d N (k) = v02 (R + 1) W (k) + v02 + v0 (v0 + k)R + kR2 + R(v0 + k)(R + v0 ). dk dk We will first prove that d dk N (k) for k = k0 is negative. We will then prove that 33 (29) d dk N (k) is k=2 k=4 k=6 k=8 k = 10 Figure 8: The numerator of (λ0 + λ1 )′ in Region 3, for k = 2, 4, 6, 8, 10. decreasing in k, for all k > k0 , hence for all k > k0 , d dk N (k) < 0, and since N (k0 ) = 0 it follows that N (k) < 0 for all k > k0 . Recall that for k = k0 : v0 = u0 − k0 , and R = R(v0 ) = R(u0 − k0 ) = of Lemma 29, d dk W [A] k=k 0 = 1 u0 .  k0 −1 u0 − 1. Also, by Part 3 Substituting all this in (29) gives k0 (u0 −k0 )2 2 +(u0 −k0 )2 +(u0 −k0 )u0 u0 This equals k0 u0 2     2    k0 k0 k0 k0 − 1 +k0 −1 + − 1 u0 − 1 + u0 − k0 . u0 u0 u0 u0 k0 +   k0 − 1 (2k0 − u0 + u20 − u0 k0 ), u0 which equals    k02 k0 . −1 (u0 − 1)(u0 − k0 ) + u0 u0 (30) Since for all k > 0, we have u − k > 1, then the first factor in (30) is negative, and the second factor is positive in Region 3, proving that indeed We now prove that d dk N (k) d dk N (k) k=k0 < 0 in Region 3. is decreasing in k, for all k > k0 . By (29) d d N (k) = v02 (R + 1) W (k) + v02 + v0 (v0 + k)R + kR2 + R(v0 + k)(R + v0 ). dk dk 34 This equals v02 (R + 1) d W [A] + 2R(R + v)k + v02 (R + 1) + v0 R(R + v0 ). dk (31) d W [A] is decreasing in k, since by R1 and R2, R + 1 > 0 in Region 3 The first term v02 (R + 1) dk and by the second statement in Lemma 29, d dk W [A] is decreasing in k. The reminder expression 2R(R + v)k + v02 (R + 1) + v0 R(R + v0 ) is a linear function of k with a negative multiplier 2R(R + v) for k, hence it is also decreasing in k. So indeed, we have proven that d dk N (k) k=k0 d dk N (k) is decreasing in k, for all k > k0 . Since < 0, then it follows that for all k > k0 , d dk N (k) < 0. Now, since N (k0 ) = 0, then it follows that N (k) < 0, for all k > k0 . This means that the numerator of (λ0 + λ1 )′ (where the derivative is by v0 ), is negative for all k > k0 . Since u − k = ln u 1 1− u is increasing in k, then for all k > 0 : u − k > u0 − k0 = v0 . Hence, for all k > 0, and for all 1 < v0 < u − k, (i.e., Region 3) (λ0 + λ1 )′ < 0, implying that λ0 + λ1 decreases in v0 in Region 3. 8.3 8.3.1 Results relating to Section 5.2 - Profit maximization Proof of Proposition 13 −λ1 1. Recall that in Region 2, λ0 = 0. Hence by (6), (v − p) 1−eλ1 p=v− = 1, and so λ1 , 1 − e−λ1 π = v(1 − e−λ1 ) − λ1 , and d π = ve−λ1 − 1. dλ1 Because d dλ1 π (32) = 0 for λ∗1 = ln v, the local maximum profit of Region 2, and the profit associated with it, are defined in (8). 2. Recall that in Region 4, λ1 = 0. By (5) we have p=v−k− λ0 . 1 − e−λ0 So by (7) π = (v − k)(1 − e−λ0 ) − λ0 . 35 Now, d π = (v − k)e−λ0 − 1, dλ0 (33) giving the profit-maximizing value λ∗0 = ln (v − k). Hence, the local maximum profit of Region 4, and the profit associated with it, are given by (9). 3. Because p3 = v − k − 1 separates between Regions 2 and 3, at this point λ0 = 0, and so
π3 = (v − k − 1) 1 − e−λ1 . Now, p = v − k − 1 implies that v0 = 1. Hence by Proposition 10, in Region 3, λ1 = W [A] − (v0 + k)R(v0 ) . v0 Thus by R2, we obtain for v0 = 1 λ1 = W [A(1)] − (k + 1)(−1) = W [−(k + 1)e−(k+1) ] + k + 1 = W + k + 1. Substituting λ1 = W + k + 1 in π3 gives (10). 8.3.2 Proof of Theorem 14 1. By (9), π4 = v − k − 1 − ln (v − k) , and by (8), π2 = v − 1 − ln (v). Note that for y ≥ 1, the value of y − 1 − ln y is increasing. That result and the inequalities 1 ≤ v − k < v imply that π4 < π2 . 2. To prove that π3 ≤ π2 , we must prove that   (v − k − 1) 1 − e−(W +k+1) ≤ v − 1 − ln v, which is equivalent to proving that   (v − k − 1) 1 − e−(W +k+1) − v + ln v ≤ −1. (34) We will find the maximum value of the left-hand side of (34) and show that it equals −1. To find the maximum we solve    ′ 1 (v − k − 1) 1 − e−(W +k+1) − v + ln v = 1 − e−W −k−1 − 1 + = 0. v 36 The solution is v = eW +k+1 , for which the second derivative is − v12 . Hence eW +k+1 is a local maximum. k+1 −(k + 1)e−k−1 ek+1 = . W −W eW +k+1 = eW ek+1 = (35) We now show that for v = eW +k+1 , the left-hand side of (34) is −1. By W5 we obtain =  !    k+1 k+1 W e−(k+1) k+1 − −k−1 + ln 1+ + W W −W (k + 1)e−(k+1) =  k+1 − −k−1 W     W k+1 k+1 1+ + ln + , k+1 W −W which gives  k+1 − k − 2 − W + ln − W Now, since by (35) k+1 −W  . (36) = eW +k+1 , then ln  k+1 −W  = ln e(W +k+1) = W + k + 1. Substituting this in (36) gives −k − 2 − W + ln 8.3.3  k+1 −W  = −k − 2 − W + W + k + 1 = −1. Proof of Lemma 16 We will first prove that p2 belongs to Region 2, iff ln v 1− v1 ≤ k+1. Then, we will prove that ln v 1− v1 ≤ k+1, iff v ≤ eW +k+1 . Recall that Region 2 refers to all p satisfying v − k − 1 ≤ p ≤ v − 1. First, by L’Hôpital’s rule limv→1 v1 ln v 1 = = = 1. 1 1 v→1 1 − 1 limv→1 v2 v lim Because ln v 1− v1 is increasing for all v ≥ 1, ln v ≥ 1, 1 − v1 and so v− ln v ≤ v − 1, 1 − v1 37 proving p2 ≤ v − 1, always. We now prove that p2 ≥ v − k − 1, iff ln v 1− v1 ≤ k + 1. The latter is equivalent to v− ln v ≥ v − k − 1. 1 − v1 Hence in this case, p2 ≥ v − k − 1, and so p2 is in Region 2. Now, we will show that the unique solution for ln v = k + 1, 1 − v1 (37) is e(W +k+1) . By (35) eW +k+1 = − k+1 . W Hence W eW +k+1 = −(k + 1), implying that (W + k + 1)eW +k+1 − (k + 1)eW +k+1 = −(k + 1), and so   (W + k + 1)eW +k+1 = (k + 1) eW +k+1 − 1 . This is equivalent to (W + k + 1)eW +k+1 = k + 1, eW +k+1 − 1 which implies W +k+1 = k + 1. 1 1 − eW +k+1 Hence eW +k+1 solves (37). Because the left-hand side of (37) is strictly increasing, eW +k+1 uniquely solves (37). 8.3.4 Proof of Lemma 19 First, note that f (k + 1) = 0. Now, ′ f (v) =  v 1− k+1  W − ln (v − k) ′ =− and we have f ′′ (v) = 1 > 0, (v − k)2 38 1 W − , k+1 v−k (38) hence f is indeed strictly concave, thus has at most two roots. To see that k +1 is not the only root, we need to find vm the minimum of f in Region 3, and show that k + 1 6= vm . To solve f ′ (v) = 0, we utilize (38), and get − W 1 − = 0. k+1 v−k This is equivalent to v−k =− k+1 . W Hence v=k− and so vm = k − k+1 W , k+1 , W is the minimum of f in Region 3. We now show that k + 1 < vm . Since −(k + 1)e−(k+1) ≥ −e−1 for all k > 0, and W (·) is an increasing function, then W = W [−(k + 1)e−(k+1) ] ≥ W [−e−1 ] = −1. Thus for all k > 0, W > −(k + 1), and so 1<− k+1 , W and k+1<k− k+1 = vm . W Hence f has exactly two roots. Since f is strictly convex and f (vm ) < 0, (since f (k + 1) = 0, and vm is the minimum of f ), then vm lies between the two roots. Since we have established that k + 1 < vm , then k + 1 < vm ≤ v f . 8.3.5 Proof of Corollary 21 Recall that by (35) we have eW +k+1 = − (k + 1) . W Thus eW +k+1 = − (k + 1) (k + 1) ≤k− = vm ≤ vf , W W where the last inequality follows from (39). 8.3.6 Proof of Lemma 23 39 (39) 1. Recall that Region 4 refers to all p satisfying 0 ≤ p ≤ v − u. To prove that p4 ≥ 0, we must prove that v−k− ln (v − k) ≥ 0. 1 1 − v−k This is equivalent to  1 (v − k) 1 − v−k  − ln(v − k) ≥ 0. Hence we need to prove that (v − k) − ln(v − k) ≥ 1. (40) Note that v − k ≥ u ≥ 1, (where the last inequality follows from the fact that u = u(k) ≥ 1, for all k ≥ 0). For v − k ≥ 1, (40) always holds since the left-hand side of (40) equals 1, for v − k = 1 and is increasing for v − k ≥ 1. Hence p4 is indeed non-negative. We now prove that p4 ≤ v − u if v ≥ k + u. Since ln y 1− y1 is increasing in y, then v − k ≥ u implies ln(v − k) ln u ≥ . 1 1 − v−k 1 − u1 Hence v−k− ln(v − k) ln u ≤v−k− . 1 1 − v−k 1 − u1 The left-hand side of the above equals p4 , so we have p4 ≤ v − k − Recall that u = k + ln u 1 1− u ln u . 1 − u1 , so that p4 ≤ v − u. 2. We will prove that k + u ≤ vm , where vm was defined as the minimum of f. This result and the observation that vm ≤ vf (see (39) in the proof of Lemma 19), complete the proof. Recall that vm = k − (k + 1) . W Hence we need to prove that u≤− 40 (k + 1) . W 15 10 5 0.5 1.0 1.5 2.0 k Figure 9: u (solid line) and eW +k+1 (dashed line) as functions of k By (35) − (k + 1) = eW +k+1 , W so we must prove that u ≤ eW +k+1 . Both sides of the inequality are functions of k. For k = 0, we obtain equality with 1 on both sides. 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