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.

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    $\begingroup$ The integral does not need to converge for all values of $s$. In fact, $G(s)$ only exists for $|s(1-p)|<1$. Compute the integral for all values of $S$ for which the integral exists; what do you get? $\endgroup$ – Mike Earnest Mar 19 at 0:39
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    $\begingroup$ Also, there are two flavors of geometric distribution; either the support is nonnegative integers, or positive integers. $V$ is of the first flavor, while the $G(s)$ you wrote is for the second flavor, so do not try to match $G_V(s)$ to your other $G(s)$. $\endgroup$ – Mike Earnest Mar 19 at 0:42
  • $\begingroup$ Thank you! The fact that there are two types of geometric distribution really helped me out. $\endgroup$ – Renat Sergazinov Mar 21 at 17:45

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