Timeline for Efficient Computation of convolution-like expression
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Dec 9 at 10:11 | vote | accept | tiX5eeMo | ||
| Dec 9 at 9:22 | comment | added | tiX5eeMo | Thank you for your reply. The function mu(x_1,...,x_m) is arbitrary non-negative. | |
| Dec 8 at 19:59 | comment | added | Dillon Davis |
@tiX5eeMo If mu(x_1,...,x_m) is some arbitrary function, then you'll have no choice but to evaluate it separately 2^m times (each possible x_i assignment). If you can tell us more about mu, we may be able to provide a better solution, but right now you haven't provided enough detail. An example for m=3 would help immensely.
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| Dec 8 at 19:40 | comment | added | tiX5eeMo | I think you are assuming that the distribution mu(x_1,...,x_m) is independent. But the distribution mu(x_1,...,x_m) is not necessarily independent. | |
| Dec 8 at 9:21 | history | answered | Dillon Davis | CC BY-SA 4.0 |