determine distribution by finding the moment generating function Mathematical Statistics and Data Analysis, Rice, Chapter 4, Problem 92
$\theta$ is Gamma($\lambda,\alpha$) distributed, $X|\theta$ follows a Poisson($\theta$) distribution. Wanted: the unconditional distribution of $\alpha + X$.
Attempt:
$M_{\alpha+X}(t) = e^{\alpha t}E(E(e^{tX}|\theta)) = e^{\alpha t}E(M_{X|\theta}(t)) = e^{\alpha t}E(e^{\theta(e^t-1)}) = e^{\alpha t}M_{\theta}(e^t-1) = \left(\frac{\lambda e^t}{(\lambda-e^t+1)}\right)^\alpha$
Can someone help me along? How to identify to which distribution the mgf belongs?
 A: The OP arrived at the expansion $$M_{\alpha+X}(t)=E(\mathrm e^{t(\alpha+X)})=\left(\frac{\lambda e^t}{\lambda-e^t+1}\right)^\alpha.$$ Factoring the constant factor and the factor involving a power of $\mathrm e^t$, and using the expansion $$\frac1{(1-u)^\alpha}=\sum_{k\geqslant0}{\alpha+k-1\choose k}u^k,$$ for the argument $$u=\frac{\mathrm e^t}{\lambda+1},$$ this is also
$$
M_{\alpha+X}(t)=\left(\frac{\lambda \mathrm e^t}{\lambda+1}\right)^\alpha\frac1{(1-u)^\alpha}=\left(\frac{\lambda \mathrm e^t}{\lambda+1}\right)^\alpha\sum_{k\geqslant0}{\alpha+k-1\choose k}\left(\frac{\mathrm e^t}{\lambda+1}\right)^k.$$
Hence
$$\sum_{k\geqslant0}\mathrm e^{t(\alpha+k)}P(X=k)=\left(\frac{\lambda}{\lambda+1}\right)^\alpha\sum_{k\geqslant0}{\alpha+k-1\choose k}\left(\frac1{\lambda+1}\right)^k\mathrm e^{t(\alpha+k)},$$
and, by identification, for every $k\geqslant0$,
$$P(X=k)=P(\alpha+X=\alpha+k)=\left(\frac{\lambda}{\lambda+1}\right)^\alpha{\alpha+k-1\choose k}\left(\frac1{\lambda+1}\right)^k.$$ Thus, the distribution of $X$ is negative binomial $(\alpha;p)$, with $$p=\frac1{\lambda+1}.$$
