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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?

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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.

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  • $\begingroup$ Genius! I wish I could see these things. Thank you so much! $\endgroup$ – ereHsaWyhsipS Feb 5 '18 at 0:37
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    $\begingroup$ @ereHsaWyhsipS You're welcome! I am flattered and suppose my answer makes it seem like I magically identified the MGF but I did not...I just knew from the context of the given Poisson-Gamma mixture that you'd end up with a Negative Binomial distribution. So, I looked up its MGF and confirmed the match with the substitution--after a few failed attempts. $\endgroup$ – Nap D. Lover Feb 5 '18 at 0:56

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