Let $N$ by the number of claims made by a person. Assume that the Number of claims varies with the type of person. Measure the type of person by a second random variable $\Lambda$, which is nonnegative and follows some distribution $F$. Further, assume $N$ conditioned on $\Lambda =\lambda$ follows a Poisson with mean $\lambda$.
Find the distribution of $N$ using the MGF technique; that is, express it as a function of the MGF of $\Lambda$.
Find the exact expression of the MGF of $N$ if $\Lambda$~$GAMMA(\alpha, \beta)$, and conclude the distribution of $N$ using the uniqueness theorem.
So here's where I've made it so far:
$$M_N(t)=E(e^{tN})=E(E(e^{tN}|\Lambda))=E(e^{\Lambda(e^t-1)})=M_\Lambda(e^t-1)$$
So if if $\Lambda$~$GAMMA(\alpha, \beta)$, we have that $$M_N(t)=\frac{1}{(1-\beta e^t+\beta)^\alpha}$$
But I do not recognize this MGF, which suggests that I didn't do something correctly somewhere. Do you recognize this MGF? If not, what is my mistake?
Best Answer
If $M_{\Lambda}(t)=(1-t/\beta)^{-\alpha}$ is the MGF of your Gamma variable, then $$M_N(t)=M_{\Lambda}(e^t-1)=\left(\frac{\beta}{\beta+1-e^t}\right)^{\alpha}$$ $$=\left(\frac{q}{1-pe^t}\right)^{r},$$ using the substitution $\beta=q/p$, $q=1-p$ and $r=\alpha$. We note that this is the MGF of a Negative Binomial RV.