# Poisson Variable with an Exponential Parameter becoming a Geometric Distribution?

Suppose Λ ∼ exponential(γ) and X ∼ Poisson(Λ). Use moment generating functions to show that $$X + 1 \sim \mathrm{geometric}(p)$$ and determine $$p$$ in terms of γ.

In order to solve this problem, I first did:

$$E[e^{s(X+1)}] = e^sE[e^{sX}]$$

Then I found $$E[e^{sX}]$$ by using iterated expectation:

$$E[e^{sX}] = E[E[e^{sX}| Λ = y]] = \int_{0}^{\infty}e^{y(e^s - 1)}γe^{γy}$$

Once simplified, I got:

$$E[e^{s(X+1)}] = \frac{γe^s}{e^s - 1 - γ}$$

However, the MGF of a Geometric Variable is $$\frac{pe^s}{1-(1-p)e^s}$$, and I can't seem to find a way to match the two equations and find how to determine p in terms of γ.

Recheck your calculation of the MGF of the exponential distribution; there is a mistake of sign.

As you have done,

\begin{align} E\left[e^{s(X+1)}\right]&=E\left[E\left[e^{s(X+1)}\mid \Lambda\right]\right] \\&=e^sE\left[\exp\left(\Lambda \left(e^{s}-1\right)\right)\right] \\&=e^s \left[\frac{\gamma}{\gamma-(e^s-1)}\right]\qquad,\, s<\ln(\gamma+1) \\&=\frac{\gamma e^s}{1+\gamma-e^s} \\&=\frac{\left(\frac{\gamma}{\gamma+1}\right)e^s}{1-\left(\frac{1}{\gamma+1}\right)e^s} \end{align}

Now it is clear that $$X+1$$ is geometric.