Timeline for Reversing the order of conditioning in a sum to compare conditional variances
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 16, 2020 at 18:40 | vote | accept | PeteJorgensen | ||
| Jun 8, 2020 at 8:24 | comment | added | ofer zeitouni | @PeteJorgensen The argument for the general case (ie when the density of $X$ does not have a bounded Fisher information) escapes me. | |
| Jun 8, 2020 at 8:23 | comment | added | ofer zeitouni | @PeteJorgensen Now, there is a lower bound on $\sigma^2$, known as the van-Trees version of the Cramer Rao lower bound from statistics, which gives a lower bound on $\sigma^2$ in terms of the inverse of a sum of two Fisher informations: one related to $p(Z|X)$ (which involves the Gaussianity of $Y$, and hence is bounded), and the other is the Fisher information of the law of $X$. If the latter is bounded (e.g., if $X$ has a nice smooth density that does not vanish too fast at a point) then the latter is bounded and you get a lower bound on $\sigma^2. | |
| Jun 8, 2020 at 8:22 | comment | added | ofer zeitouni | @PeteJorgensen Yes. The question you are asking can be rephrased as a question about the mean square error of the optimal estimator of $X$ given $Z$. Namely, your conjecture is equivalent to the claim that $\sigma^2:=\inf_{W} E(X-W)^2>0$, where the infimum is over all $Z$-measurable, $L^2$ random variables (the infimum is achieved by $W=E(X|Z)$, and this is the link to your question). $\sigma^2$ is the variance I refer to in my answer. | |
| Jun 7, 2020 at 18:33 | comment | added | PeteJorgensen | @oferzeitouni Can you elaborate on this part: "...would give a strictly positive lower bound on the variance. This however does not prove the general case."? Which variance and what kind of lower bound? I think my confusion is purely semantic, i.e. there are several "variances" in the question. Apologies if this is not clear! | |
| Jun 6, 2020 at 11:27 | comment | added | ofer zeitouni | This is a nice question. The issue can be reformulated as finding a lower bound on the estimation error of $X$ given $Z$. Now, if the law of $X$ has finite fisher information, then the Gaussianity of $Y$ together with Van-Tree's version of Cramer-Rao would give a strictly positive lower bound on the variance. This however does not prove the general case. | |
| Jun 3, 2020 at 23:27 | comment | added | Nate Eldredge | @S.Surace: It's definitely true that if $\operatorname{Var}[X \mid Z]=0$ then $X$ is $Z$-measurable, i.e. $X =f(Z)$. The last part I have to think about. | |
| Jun 3, 2020 at 23:26 | comment | added | S.Surace | @NateEldredge Essentially that is what I use in my answer. Maybe you can help me make the second part completely bulletproof. | |
| Jun 3, 2020 at 23:13 | answer | added | S.Surace | timeline score: 3 | |
| Jun 3, 2020 at 20:07 | review | First posts | |||
| Jun 3, 2020 at 20:41 | |||||
| Jun 3, 2020 at 20:03 | history | asked | PeteJorgensen | CC BY-SA 4.0 |