Timeline for Binomial distribution conditional on the weigthed sum?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29, 2023 at 20:29 | comment | added | Henry | Suppose $n=2 $ and the weights are $w_1=1$ and $w_2=10$: then the possible weighted sums $w^Tx$ are $0$ (in which case the unweighted sum is $0$ with conditional variance $0$), or $1$ (in which case the unweighted sum is $1$ with conditional variance $0$), or $10$ (in which case the unweighted sum is $1$ with conditional variance $0$), or $11$ (in which case the unweighted sum is $2$ with conditional variance $0$) | |
| May 29, 2023 at 19:41 | comment | added | entropy | Variance of the LHS is just p(1-p) because it is simply a binomial distribution. In the limiting case the two variances are just equal to each other. I do not see how you concluded that rhs will be zero. I do not think so. | |
| May 29, 2023 at 9:09 | comment | added | Henry | The conditional probability distribution of the unweighted sum is going to depend on the actual weights and their relationship: there will be weights where the weighted sum fully determines the unweighted sum and so there is $0$ conditional variance, such as when $n=2$ and the weights are not equal. | |
| S May 28, 2023 at 21:59 | review | First questions | |||
| May 28, 2023 at 22:01 | |||||
| S May 28, 2023 at 21:59 | history | asked | entropy | CC BY-SA 4.0 |