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There is a result that says a Negative Binomial($\mu,\phi$) distribution converges to a Poisson($\mu$) distribution when $\phi\rightarrow\infty$. Mathematically, I have:

$$f(y,\phi,\mu) = \frac{\Gamma(\phi + y)}{\Gamma(y+1)\Gamma(\phi)} \left( \frac{\mu}{\mu + \phi} \right)^y \left( \frac{\phi}{\mu +\phi} \right)^\phi, \ \ y=0,1,\dots, \ \ \phi>0, \ \ \mu\geq0$$

$$\lim_{\phi\rightarrow\infty} f(y,\phi,\mu) = \frac{\mu^y e^{-\mu}}{y!} $$

I already have $\Gamma(y+1) = y!$ and $\mu^y$, so if this holds, then I should have:

$$\lim_{\phi\rightarrow\infty} \frac{\Gamma(\phi+y)}{\Gamma(\phi)} \frac{\phi^\phi}{(\mu + \phi)^{y+\phi}} = e^{-\mu} $$

Is the previous equality true? How can you solve that?

VFreguglia
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    Perhaps this might be helpful?$$\Gamma(\phi + y) = (\phi + y - 1)\Gamma(\phi+y-1)\text{,}$$ so $$\Gamma(\phi+y) = (\phi + y - 1)(\phi + y - 2) \cdots (\phi + y - (y-1))(\phi + y-y)\Gamma(\phi+y-y) = \Gamma(\phi)\prod\limits_{k=1}^{y}\left(\phi + y - k\right)\text{,}$$giving $$\dfrac{\Gamma(\phi+y)}{\Gamma(\phi)} = \prod\limits_{k=1}^{y}\left(\phi + y - k\right)\text{.}$$ – Clarinetist Jun 23 '15 at 21:42

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