The joint distribution of X and Y is given by $f(x,y)=\frac{\exp(−y)}{y}$
where $0 So $f_Y(y) = \int_0^y \frac{\exp(−y)}{y} \mathrm{d}x = \exp(-y)$ which makes $f_{X|Y}(x,y) = \frac{f(x,y)}{f_Y(y)}= \frac{\exp(-y)/y}{\exp(-y)} = \frac{1}{y}$ so $\mathbb{E}(X^2+Y^2 |Y =y) = \int_{x=0}^y (x^2+y^2)\frac{f(x,y)}{f_Y(y)} \mathrm{d}x = \int_{x=0}^y \frac{(x^2+y^2)}{y}\mathrm{d}x = 2y^2$ Is this correct? I'm a bit confused?!
Linear conditional expectation
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$\begingroup$
probability
1 Answers
2
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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0Yes - turns out I can't integrate anymore!! – 2012-09-23