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As it was proved in the answer here, for $Z_1, Z_2$, two independent and identically distributed random variables. Then we have: $$ \mathbb E[Z_1\mid Z_1+Z_2] =\frac{Z_1+Z_2}{2}. $$

However, suppose now that $Z_1, Z_2$ are two standard normal random variables and that we define $W_1 = aZ_1 \sim \mathcal N(0, a^2)$ and $W_2 = bZ_2 \sim \mathcal N(0, b^2)$.

Thus, $W_1, W_2$ are two still independent normal random variables, but with different distributions as stated. Question: what is $$ \mathbb E[Z_1\mid W_1+W_2]? $$

The problem is that now the argument in the question linked seems not to be working here, because now $W_1+W_2$ is not anymore a symmetric function of $(Z_1, Z_2)$.

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RandomGuy
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    Indeed the symmetry argument does not work anymore. But in the normal case, one knows that $$Z_1=c(W_1+W_2)+dW+e$$ for some constants $(c,d,e)$ and $W$ standard normal independent of $W_1+W_2$, hence $$E(Z_1\mid W_1+W_2)=c(W_1+W_2)+e.$$ In your case, everybody is centered hence $e=0$. Furthermore, $E(Z_1(W_1+W_2))=cE((W_1+W_2)^2)$, that is, $a=c(a^2+b^2)$ and $$E(Z_1\mid W_1+W_2)=\frac{a}{a^2+b^2},(W_1+W_2).$$ – Did Jun 30 '16 at 12:13
  • See also this question: http://math.stackexchange.com/questions/1368922/conditional-expectation-on-gaussian-random-variables/1368986#1368986. – Gordon Jun 30 '16 at 15:35

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