Find the marginal density of X given that (fx|y=y(x)) = e^(-y)*y^x/x!
(i.e., that the conditional density of X, given that Y=y is Poisson(y))
and that fy(y) = lamda^a / gamma(a) * y^(a-1)*e^(-lamda*y) where a(alpha) is a positive integer (i.e., the random variable Y has gamma(a,lamda) density). Hint: fx(x) is a well-known density.

???? ㅡㅡ;;

2. Originally Posted by ninano1205
Find the marginal density of X given that (fx|y=y(x)) = e^(-y)*y^x/x!
(i.e., that the conditional density of X, given that Y=y is Poisson(y))
and that fy(y) = lamda^a / gamma(a) * y^(a-1)*e^(-lamda*y) where a(alpha) is a positive integer (i.e., the random variable Y has gamma(a,lamda) density). Hint: fx(x) is a well-known density.

???? ㅡㅡ;;
The starting point would be

$f_X(x | y) = \frac{f(x, y)}{f_Y(y)} \Rightarrow f(x, y) = f_X(x | y) \, f_Y(y)$

where f(x, y) is the joint pdf of X and Y.

Now substitute the given distributions for $f_X(x | y)$ and $f_Y(y)$.

Now integrate f(x, y) with respect to y to get $f_X(x)$.

3. But the real problem is the integration.
I cannot proceed from the ingral of f(x,y) with repect to y.
Anythought?

4. Originally Posted by dingdong
But the real problem is the integration.
I cannot proceed from the ingral of f(x,y) with repect to y.
Anythought?
After substituting the various pdf's and simplifying, you have to deal with the following integral:

$\int_{0}^{+\infty} y^{\alpha + x - 1} e^{-(\lambda + 1) y} \, dy$.

I suggest making the substitution $t = (\lambda + 1) y$ to get the integral representation of the gamma function.

Note: $\int_{0}^{+\infty} t^{\alpha + x - 1} e^{-t} \, dt = \Gamma (\alpha + x)$.