I am doing a review for the test, and I have found myself really struggling with the following question:
Prove using generating functions that if $$V \sim \mathrm{Poi}(\theta),~\theta \sim \mathrm{Exp}(\nu),$$ then $V$ has geometric distribution.
My attempt: $$G_V(s) = E\left[ s^V \right] = E\Big[ E\left[ s^V \middle|\, \theta \right] \Big] = \int_0^\infty{ E\left[ s^V \middle|\, \theta = x \right] f_\theta(x)dx}$$ $$E\left[ s^V \middle|\, \theta = x \right] = \sum_0^\infty{ s^k \frac{ x^k e^{-x} }{k!} }=e^{-x} e^{sx} = e^{x(s-1)}$$ Hence plugging back in and substituting pdf of $\theta$, I get:$$\int_0^\infty{ e^{x(s-1)}\nu e^{-\nu x}dx}$$ Now I am stuck since the integral does not even converge for some values of $s$. I know that generating function of geometric distribution is:$$G(s)=\frac{ps}{1-(1-p)s},$$but I don’t know to get there from what I have.