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The joint distribution of X and Y is given by $f(x,y)=\frac{\exp(−y)}{y}$ where $0<x<y<\infty$. Compute $\mathbb{E}(X^2+Y^2 |Y =y)$.

So $f_Y(y) = \int_0^y \frac{\exp(−y)}{y} \mathrm{d}x = \exp(-y)$

which makes

$f_{X|Y}(x,y) = \frac{f(x,y)}{f_Y(y)}= \frac{\exp(-y)/y}{\exp(-y)} = \frac{1}{y}$


$\mathbb{E}(X^2+Y^2 |Y =y) = \int_{x=0}^y (x^2+y^2)\frac{f(x,y)}{f_Y(y)} \mathrm{d}x = \int_{x=0}^y \frac{(x^2+y^2)}{y}\mathrm{d}x = 2y^2$

Is this correct? I'm a bit confused?!

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1 Answer 1

up vote 1 down vote accepted

Everything looks good to me except

$\int_{x=0}^y \frac{(x^2+y^2)}{y}\mathrm{d}x = \frac{4y^2}{3}$ instead of $2y^2$

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Yes - turns out I can't integrate anymore!! –  John Peters Sep 23 '12 at 17:56

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