Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?!

share|cite|improve this question

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 $2y^2$

share|cite|improve this answer
Yes - turns out I can't integrate anymore!! – John Peters Sep 23 '12 at 17:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.