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$X$ and $Y$ are two random variables drawn independently from standard normal distribution $\mathcal{N}(0,1)$. Set $Z:=3X+4Y$. What is the conditional expectation of $X$ given $Z$, i.e., $E[X|Z]?$

By symmetry, I guess $E[X|Z] = \frac{1}{7}Z$. Not sure if it is correct..

More generally, if $X \sim\mathcal{N}(0,\sigma_1^2)$ and $Y \sim\mathcal{N}(0,\sigma_2^2)$ are independent, any ideas how to calculate $E[X|X+Y]$?

zxzx179
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    Find a linear combination of $X$ and $Y$ - call it $Z$ - that is uncorrelated with (and hence independent of) $H=X+Y$. Express $X$ as a linear combination of $H$ and $Z$ and hence obtain $E(X|Z)$. – A.S. Dec 06 '15 at 04:53
  • You get $E(X|X+Y)=\frac {\sigma_1^2}{\sigma_1^2+\sigma_2^2}(X+Y)$. – A.S. Dec 06 '15 at 05:00
  • See also discussion at http://math.stackexchange.com/questions/1368922/conditional-expectation-on-gaussian-random-variables/1368986#1368986. – Gordon Dec 18 '15 at 18:07

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