This question is Exercise 3.15 in Statistical Inference by Casella and Berger. It asks to prove that the MGF of a Negative Binomial $\mathscr{N}eg(r,p)$ converges to the MGF of a Poisson $\mathscr{P}(\lambda)$ distribution, when
$$r\to\infty\,, \quad p\to 1\,, \quad r(1-p)\to\lambda$$
The formula I have for the MGF of $X\sim \mathscr{N}eg(r,p)$ is:
$$M_X(t) = \frac{p^r}{[1-e^t(1-p)]^r}$$
Considering just the denominator, we have
$$[1-e^t(1-p)]^r = [1+\frac{1}{r}e^tr(p-1)]^r = [1+\frac{1}{r}e^t(-\lambda)]^r$$ As $r\to\infty$, this converges to $e^{-\lambda e^t}$. Now considering the entire formula again, and letting $r\to\infty$ and $p\to 1$, we get $e^{\lambda e^t}$, which is incorrect since the MGF of Poisson($\lambda$) is $e^{\lambda(e^t-1)}$. I seem to be on the right track, just made a misstep somewhere. Can anyone spot my mistake?
Best Answer
You make a mistake by ignoring $p^r$: If you consider your MGF $$M_X(t) = \frac{p^r}{[1-e^t(1-p)]^r}\,,$$ then $$\log\{M_X(t)\} = r\log(p)-r\log\{1-e^t(1-p)\}$$ and using the asymptotic equivalences \begin{align*} r\log(p)-r\log\{1-e^t(1-p)\} &= r\log(1-[1-p])-r\log\{1-e^t(1-p)\}\\ &\approx -r[1-p]+re^t(1-p)\\ &\approx \lambda[-1+e^t] \end{align*} which shows that the limiting value of the MGF is $$ \exp\{\lambda[e^t-1]\}$$ as requested in this exercise.