What is the distribution ?

Question

Q===================================================================

$f(Θ)$ is pdf of gamma distribution

$$f(Θ) = \frac{λ^α}{Γ(α)}Θ^{α-1}\exp(-λΘ), $$

$$X\mid Θ \sim \mathrm{poisson}(Θ) \rightarrow \frac{Θ^x\exp(-Θ)}{x!}$$

Suppose that $Θ$ is a random variable that follows a gamma distribution with parameters $λ$ and $α$, where $α$ is an integer, and suppose that, conditional on $Θ$, $X$ follows a Poisson distribution with parameter $Θ$. Find the unconditional distribution of $α + X$ (Hint : Find the mgf by using iterated conditional expectations.)

====================================================================

I got

$$M_{a+X} = \frac{(λexp(t))^α}{(λ-exp(t)+1)^α}$$

But I don't know distribution has this mgf.

What is the distribution ?

Answer 1

It seems easier to me, at least, to find the unconditional distribution directly: $$\begin{align*} \Pr[X = x] &= \int_{\theta = 0}^\infty \Pr[X = x \mid \Theta = \theta] f_\Theta(\theta) \, d\theta \\ &= \int_{\theta = 0}^\infty e^{-\theta} \frac{\theta^x}{x!} \frac{\lambda^\alpha \theta^{\alpha-1} e^{-\lambda \theta}}{\Gamma(\alpha)} \, d\theta \\ &= \int_{\theta = 0}^\infty \frac{\lambda^\alpha \theta^{x+\alpha-1} e^{-(\lambda+1)\theta}}{x! (\alpha-1)!} \, d\theta \\ &= \frac{(x+\alpha-1)!}{x! (\alpha-1)!} \frac{\lambda^\alpha}{(\lambda+1)^{x+\alpha}}\int_{\theta=0}^\infty \frac{(\lambda+1)^{x+\alpha} \theta^{x+\alpha-1} e^{-(\lambda+1)\theta}}{\Gamma(x+\alpha)} \, d\theta \end{align*}$$ and then recognizing that the integrand is the density of a gamma distribution with rate $\lambda + 1$ and shape $x+\alpha$, thus equals $1$. It follows that $$\Pr[X = x] = \binom{x+\alpha-1}{\alpha-1} \left(\frac{1}{\lambda+1}\right)^x \left(\frac{\lambda}{\lambda+1}\right)^\alpha$$ which is quite clearly negative binomial. I leave it to you to determine what the relevant parameters are; then $\alpha + X$ is also negative binomial, albeit with a different support and parametrization.

---

For the sake of completeness, we will do the calculation of the moment generating function. Recall that the MGFs of the Poisson and gamma distributions are, respectively $$M_{X \mid \Theta}(t) = \operatorname{E}[e^{tX}\mid\Theta] = \exp((e^t-1)\Theta),$$ and $$M_\Theta(m) = \operatorname{E}[e^{m\Theta}] = (1 - m/\lambda)^{-\alpha}$$ for the chosen shape and rate parametrization you wrote in your question. Consequently, $$M_X(t) = \operatorname{E}[e^{tX}] = \operatorname{E}[\operatorname{E}[e^{tX}\mid\Theta]] = \operatorname{E}[M_{X\mid\Theta}(t)] = \operatorname{E}[\exp((e^t-1)\Theta)] = M_\Theta(e^t-1) = \left(1 - \frac{e^t-1}{\lambda}\right)^{-\alpha}.$$ This of course is equivalent to $$M_X(t) = \left(\frac{\lambda/(1+\lambda)}{1 - e^t(1 - \lambda/(1+\lambda))}\right)^\alpha,$$ and now we can see that $p = \lambda/(1+\lambda)$ will give us the form of a negative binomial MGF.