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Let $X$ and $Y$ be random variables such that $Y|X = x$ is $\text{Exp}(1/x)$ and X is $\text{Gamma}(2,1)$ find the pdf of $Y$ and $E(Y)$

meta_warrior
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jack
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    Hi Jack, this looks like a homework question. Can you show us what you have done so far? – meta_warrior Dec 14 '13 at 06:31
  • The standard procedure works. Where is the trouble? – Did Dec 14 '13 at 08:18
  • I calculate P(Y=y)= integral from 0 to infinity P(Y=y|X=x)* fx(x)dx = integral from 0 to infinity xe^(-yx)1/(Gamma(2))x * e^(-x) dx – jack Dec 15 '13 at 06:27
  • I can't integral this function. For expect value. I use E(Y|X=x)= integral from 0 to infinity yxe^(-x^2)dy. I still can't figure. I know once I get it. I will try E(E(Y|X)), because it is equal to E(Y) – jack Dec 15 '13 at 06:38

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Hint:

$f_{Y|X}(y|x)f_X(x)=f(x,y)\\ f_Y(y)=\int_{-\infty}^{\infty} f(x,y)dx$

hhsaffar
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  • integral from 0 to infinity xe^(-yx)1/(Gam(2) x * e^(-x) dx to get xe^(-yx)1/(Gamma(2)) x * e^(-x) dx. How do I integral it? is this correct? – jack Dec 15 '13 at 06:41