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    Expected Value question.




    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.

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    Re: Expected Value question.

    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...

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    Re: Expected Value question.


    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?

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