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What is the conditional expectation $E(x\mid x+y)$ where independent $x,y$ are normally distributed as $x,y\sim \mathcal{N}(\mu,\sigma^2)?$

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I'm going to call them capital $X$ and capital $Y$ so that lower-case $x$ and $y$ will be available for such expressions as $\Pr(X\le x)$ or $f_X(x)$, etc., even though I won't use those here, simply because attention to such things is too often neglected.

I'm also going to assume for now that $X$ and $Y$ are jointly normally distributed, although that was not stated. Let $\rho$ be the correlation between $X$ and $Y$. The we have $$ \begin{bmatrix} X \\ Y \end{bmatrix} \sim \mathcal N\left( \begin{bmatrix} \mu \\ \mu \end{bmatrix}, \sigma^2 \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix} \right). $$ Let $$ \begin{bmatrix} U \\ V \end{bmatrix} = \begin{bmatrix} X-Y \\ X+Y \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} X \\ Y \end{bmatrix} = A \begin{bmatrix} X \\ Y \end{bmatrix}. $$ Then $$ \begin{bmatrix} U \\ V \end{bmatrix} \sim \mathcal N \left( A\begin{bmatrix} \mu \\ \mu \end{bmatrix}, (1-\rho)A\begin{bmatrix} \sigma^2 & 0 \\ 0 & \sigma^2 \end{bmatrix}A^T \right) = \mathcal N \left( \begin{bmatrix} 0 \\ 2\mu \end{bmatrix}, 2\sigma^2\begin{bmatrix} 1-\rho & 0 \\ 0 & 1+\rho \end{bmatrix} \right). $$ Hence $X-Y$ is independent of $X+Y$. Therefore $(X-Y)\mid(X+Y) \sim \mathcal N (0, 2\sigma^2(1-\rho))$. Therefore $$ \frac{X-Y} 2 \mid (X+Y) \sim \mathcal N\left( 0, \frac{\sigma^2} 2 (1-\rho) \right). $$ When conditioning on $X+Y$, the act of adding $(X+Y)/2$ to a random variable amounts to adding a constant. Hence $$ \frac{X-Y} 2 + \frac {X+Y} 2 \mid (X+Y) \sim \mathcal N\left( \frac{X+Y} 2, \frac{\sigma^2} 2 (1-\rho) \right). $$ In other words, $$ X\mid (X+Y) \sim \mathcal N\left( \frac{X+Y} 2, \frac{\sigma^2} 2(1-\rho) \right). $$

There is a short argument from symmetry for the conditional expected value, but that doesn't give the conditional variance, and the technique used here is sometimes useful elsewhere.

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    $\begingroup$ Nice approach +1. But what I meant in my comment above: $(X,X+Y)^T = A (X,Y)^T$ where $A=[1, 0; 1,1]$, as such we readily see the joint dist of $X$ and $X+Y$ which lends to a direct solution. $\endgroup$ – Math-fun Jul 7 '16 at 7:47
  • $\begingroup$ @Math-fun : But it seems simpler to do it without explicitly dealing with the density function. $\qquad$ $\endgroup$ – Michael Hardy Jul 7 '16 at 19:27

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