# Expected Value question.

#### c2q

##### New Member
Suppose that X, given Y = y, is distributed Gamma(α, y) and that the marginal distribution of Y is given by
1/Y~ Exponential(λ). Determine E(X).

For this question I'm using the theorem that states E(X) can be determined by E(E(X|Y)).

The distribution of X|Y is given by that gamma distribution listed above.
Thus E(X) can be determined by:

integral E(X|Y)Fy(y) dy

My problem is not knowing how to write Fy(y). It says the marginal distribution is written as 1/Y ~ Exp, how exactly do i write that? For example when it says

Y~Exp(lamda), that means the distribution of Y can be written as:

lamda(e)^(-lamda)(y)

so how would Fy(y) be written in this case?

Thanks.

#### c2q

##### New Member
One more note. I got a really strange answer and I'm not sure if I'm doing it right. As i said,

E(X) = integral E(X|Y=y)Fy(y)

So in order to find E(X|Y) this is what i did

E(X|Y) = integral x fx|y(x|y) dx

now i used integration by parts here, and set dv to fx|y(x|y) dx, that means that v would be 1 since that is a gamma function and would integrate to 1 wouldn't it?

in which case u = x and du = dx

After simplifying i get E(X|Y) as...0 which results in E(X) = 0 as well...

#### Dason

##### Ambassador to the humans
Question: What is the expected value of a Gamma($$\alpha, \beta$$)? Because to get E[X|Y] all you are looking for is the expected value of a Gamma(a, y). IE in your formula for the expected value of a gamma plug in a for $$\alpha$$ and y for $$\beta$$.

Also there is a very nice little formula for E[$$X^v$$] when X ~ Gamma(a,b). Have you learned anything along those lines?