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$ X $ and $ Y $ are both independent and identically distributed random variables with normal distributions $ \mathcal N ( 0 , 1 ) $. What is $ E ( X Y | X + Y = w ) $?

I know this means that $ W=X+Y $ must be normally distributed as well with mean $ 0 $ and variance $ 2 $. I also know that $ E ( X Y ) = E ( X ) E ( Y ) $ because of independence. However, I am confused as to how we calculate the conditional expectation in this situation. Do we just take the integral of the normal pdf? What would the boundaries be?

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  • $\begingroup$ Note that if $X,Y$ are i.i.d. $\mathcal{N}(0,1)$, then $W = X+Y$ is $\mathcal{N}(0,2)$. See here. $\endgroup$ – Mark Feb 24 '17 at 23:23
  • $\begingroup$ Yes, I meant to say variance of 2. However, I am confused about how to carry out the integration. $\endgroup$ – Roberto Feb 24 '17 at 23:26
  • $\begingroup$ I'm not sure, but it seems to be like that $\mathbb{E}(XY|X+Y=w) = \mathbb{E}(X(w-X)) = -\mathbb{E}(X^2) = -1$ $\endgroup$ – Andrei Kulunchakov Feb 24 '17 at 23:26
  • $\begingroup$ That's an interesting idea, but would that expectation not be zero? If the mean of X is zero, why would that be -1? $\endgroup$ – Roberto Feb 24 '17 at 23:33
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Letting $W = X + Y$ then you can write out the joint Gaussian as \begin{align*} \begin{bmatrix} X \\ Y \\ W \end{bmatrix} \sim \mathcal{N} \left( \begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}, \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix} \right) \end{align*} without worrying too much that this covariance matrix is singular, therefore partitioning and conditioning we have \begin{align*} \begin{bmatrix} X \\ Y \end{bmatrix} \bigg| W = w \sim \mathcal{N}\left( \begin{bmatrix} \frac{w}{2} \\ \frac{w}{2} , \end{bmatrix} \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix} \right) \end{align*} Still singular but that's ok because of course this Gaussian is supported on the one dimensional space $X + Y = w$. Anyway we get \begin{align*} \mathbb{E}\left[ X Y | X + Y = w \right] &= \mbox{Cov}_{X + Y = w}(X,Y) + (\mathbb{E}[X|X+Y=w])^2 \\ &= \frac{w^2}{4} - \frac{1}{2}. \end{align*}

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  • $\begingroup$ The last step (after "Anyway we get") is incorrect, for example $\frac{w^2}4-\frac12$ depends on $w$ while $E(XY)$ does not. Please explain. (As always, slightly surprising to see such answers readily accepted, in the present case after 33 minutes...) $\endgroup$ – Did Mar 8 '17 at 7:53
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Consider the rotation $$(U,V) = \left(\frac{X+Y}{\sqrt{2}}, \frac{X-Y}{\sqrt{2}}\right),$$ which by the symmetry of the bivariate standard normal, gives $(U,V) \sim \operatorname{Normal}((0,0), I_{2x2})$. Then $XY = \frac{1}{2}(U^2 - V^2)$, and under the condition $X+Y = w$, it follows that $$\operatorname{E}[XY \mid X+Y = w] = \operatorname{E}\left[\frac{U^2-V^2}{2} \biggl | \; U = \frac{w}{\sqrt{2}}\right] = \frac{w^2}{4} - \frac{\operatorname{E}[V^2]}{2} = \frac{w^2}{4} - \frac{1}{2}.$$

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