Timeline for MMSE estimator expressed through cumulants
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S May 7, 2023 at 14:15 | history | suggested | TryingToLearn | CC BY-SA 4.0 |
fixed obvious typo
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| May 5, 2023 at 11:13 | review | Suggested edits | |||
| S May 7, 2023 at 14:15 | |||||
| Oct 15, 2012 at 11:00 | answer | added | Yair Carmon | timeline score: 3 | |
| Sep 17, 2012 at 21:33 | comment | added | Pierre Robert | ok, let me rephrase my question: for $x$ distributed as a multivariate Gaussian vector, the expectation $E(X|Y)$ is linear in Y (plus a constant). I Believe that this could be seen from the fact that only the two first cumulants of a Gaussian distribution are non-zero. What I am after here is an "update" formula for the linear MMSE of non-Gaussian $X$, expressed through cumulants. | |
| Sep 7, 2012 at 14:23 | comment | added | Josh | Cumulants are statistical deterministic measures, so "sum of cumulants" can't express conditional expectation which is a "random variable". If you meant that $E(X\mid Y)$ can be represented as, for example, a power series, then I believe that Alexander's example contradict your conjecture. | |
| Sep 6, 2012 at 19:47 | comment | added | Pierre Robert | not really... Of course, the solution is clearly $E[X|y]$, and my idea is that it would be possible to express this as a sum of cumulants. But so far, I don't have any good idea on how to attack the problem. I did some searches, but found nothing. | |
| Sep 6, 2012 at 19:39 | comment | added | Alexander Chervov | MMSE estimator can ALWAYS be expressed as conditional mean E[X|Y]. Have you tried to compare yours conjecture with this ? E.g. for X_i =+-1 (BPSK modulation) ? | |
| Sep 6, 2012 at 19:20 | history | asked | Pierre Robert | CC BY-SA 3.0 |