Suppose $Y$ follows Geometric with $p$, and we know that conditional distribution of $X$ given on $Y=y$ is a $\Gamma(y,\lambda)$. What should be the conditional distribution of $Y$ given $X = x>0$?
So my attempt here is simply do: $$f_{X|Y}(x) = \frac 1 {\Gamma(y)} x^{y-1}e^{-\lambda x}$$ $$f_{Y}(y) = (1-p)^{(y-1)}p$$
such that, $$f_{Y|X}(y) = f_{X|Y}(x)f_{Y}(y) = \frac {(1-p)^{(y-1)}p} {\Gamma(y)} x^{y-1}e^{-\lambda x}$$
But I am stuck here as I am not sure if we could continue on such expression. I have tried to make it looks like a Poisson but seems it makes it worse...
$$\frac {(1-p)^{(y-1)}p} {\Gamma(y)} x^{y-1}e^{-\lambda x} = \frac{(\lambda x- \lambda xp)^{(y-1)}}{(y-1)!} e^{-(\lambda x- \lambda xp)} p \lambda e^{-\lambda xp}$$
Or do we have a better approach to deal with such questions? Thank you.