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

PT272
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    You should specify that X and Y are independent, which I suspect to be the case... – Lucien Jul 06 '16 at 12:53
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    why don't write the joint density of $X$ and $X+Y$ first? – Math-fun Jul 06 '16 at 12:53
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    @Math-fun : I'm not convinced that's the best way to approach this. $\qquad$ – Michael Hardy Jul 06 '16 at 14:21
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    See here http://math.stackexchange.com/questions/602210/compute-mathbbexxy-for-x-y-iid-standard-exponential and here http://math.stackexchange.com/questions/240471/how-to-find-exxy-k-for-geometrical-distribution and ... – leonbloy Jul 06 '16 at 14:46
  • See also questions http://math.stackexchange.com/questions/1368922/conditional-expectation-on-gaussian-random-variables/1368986#1368986 and http://math.stackexchange.com/questions/1842364/conditional-expectation-of-independent-variables/1842770#1842770. – Gordon Jul 06 '16 at 15:42
  • Thanks for the links too; I don't understand that person who requested to close this topic. This question is perfectly fine: I specified purpose "calculate conditional expectation" along with independence condition of two random variables and their distribution. That is all which is needed; just check the answer by Michael below. – PT272 Jul 07 '16 at 00:18

1 Answers1

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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.

Michael Hardy
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    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. – Math-fun Jul 07 '16 at 07:47
  • @Math-fun : But it seems simpler to do it without explicitly dealing with the density function. $\qquad$ – Michael Hardy Jul 07 '16 at 19:27