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So we have the conclusion that if $X$ and $Y$ are independent, then $g(X)$ and $h(Y)$ are independent (g, h measurable?), is it still true that for general case $X$ independent of $g(Y_1,..Y_n)$, where $Y_i$ are themselves not independent to each other.

I don't see how this is duplicate with that of showing $g(X)$ and $h(Y)$ are independent.

If it is not generally true, can we at least show however, $X$ is independent of the sum of $Y_i$?

Let me add some more condition, what about $X$ and $Y_i$ are all gaussians? And the function $g$ $h$ are nice smooth functions.

Essentially, I just want to show $X$ is independent of sum of $Y_i$.

jack
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  • I don't think it is a duplicate, since I am asking the joint, it is true for two single r.v. but how about the joint case? – jack Oct 14 '17 at 18:11
  • @user365239 How do you show that? – amsmath Oct 14 '17 at 18:22
  • @user365239 we only know $X$ is independent of each $Y_i$, $Y_i$ are not mutually independent. – jack Oct 14 '17 at 18:23
  • @user365239 it's in the question "where $Y_i$ are themselves not independent to each other." – jack Oct 14 '17 at 18:26
  • Sorry i misread – user365239 Oct 14 '17 at 18:44
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    Do you assume that $X$ is independent of $(Y_1,...,Y_n)$? –  Oct 14 '17 at 18:50
  • @d.k.o. no, that is actually exactly what I want to show. And can we at least $X$ is independent of the sum of $Y_i$? – jack Oct 14 '17 at 18:52
  • @jack The info you added significantly changed the initial question for which you got an answer. –  Oct 14 '17 at 19:01
  • Your hypothesis is unclear: is it (a) that $X$ is independent of $Y=(Y_1,\ldots,Y_n)$, or (b) that $X$ is independent of $Y_k$, for each $k$? If (a) then $X$ is trivially independent of $Y_1+\cdots+Y_n$. If (b) then $X$ may not be independent of $Y_1+\cdots+Y_n$. – Did Oct 14 '17 at 19:01
  • @Did it is (b). – jack Oct 14 '17 at 19:03
  • Then the usual (counter-)examples apply. – Did Oct 14 '17 at 19:04

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If $\sigma(X)$ is independent of $\sigma(Y_1,\ldots,Y_n)$, then $X$ is independent of any measurable function of $(Y_1,\ldots,Y_n)$. If not, here is a counterexample:

Let $Y_i=1_{A_i}$, $i=1,2$ s.t. $\mathsf{P}(A_i)=1/2$ and $A_1$ is independent of $A_2$. Then $X=1\{Y_1+Y_2 \text{ is even}\}$ is independent of $Y_i$, however, $X$ is not independent of a function of $Y_1$ and $Y_2$,

  • @N.. Compute $\mathsf{P}(X=a,Y=b)$ and compare it to $\mathsf{P}(X=a)\mathsf{P}(Y=b)$ for $a,b\in {0,1}$... –  Oct 14 '17 at 19:03
  • @N.. I'm not sure what you are asking about. –  Oct 14 '17 at 19:19