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I was reading "Maximum Likelihood Estimation for the Negative Binomial Dispersion Parameter" by Walter W. Pieogorsch, and in the intro it says the Poisson distribution is a limiting case of negative binomial distribution when the dispersion parameter $a$ goes to zero:

$$lim_{a\to0}Pr(Y=y)={ \Gamma(y+a^{-1})\over y!\Gamma(a^{-1})}({au\over 1+au})^y(1+au)^{-1/a} = {\mu^ye^{-u} \over y! }$$

I tried to work out the math, and I can see $y!$ stays the same and that $$lim_{a\to0}(1+au)^{-1/a}=e^{-u}$$

but I cannot see how $$lim_{a\to0}{ \Gamma(y+a^{-1})\over\Gamma(a^{-1})}({au\over 1+au})^y = \mu^y$$ How is this possible?

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Consider that

$$({au\over 1+au})^y=({u\over a^{-1}+u})^y$$

and then take the denominator over into the ratio of Gammas.

I think all you need to do then is make an argument that the resulting term with the gammas and the denominator goes to 1.

I believe this is one of the relations discussed in the middle of this section of the Wikipedia page on the Gamma function.

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This is covered under http://en.wikipedia.org/wiki/Negative_binomial_distribution#Poisson_distribution

The key is the parameterization of the dispersion parameter.

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  • $\begingroup$ Thank you for directing me to the link. However, I'm still confused how $\frac{\Gamma(r+k)}{\Gamma(r)(r+\lambda)^k}$ goes to 1 when r goes to infinity. $\endgroup$
    – Alby
    Commented Oct 7, 2014 at 21:45
  • $\begingroup$ Consider the simple case of factorials. $\frac{ \prod_{i=1}^k (r+i-1)...(r-1) } { \prod_{i=1}^k (r+\lambda) }$ for bounded $k$ and $\lambda$ this goes to 1 as $r$ goes to $\infty$. $\endgroup$
    – Jonathan Lisic
    Commented Oct 7, 2014 at 21:58
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    $\begingroup$ Actually, because $k$ is integral, you are in that simple case no matter what, since $\Gamma(r+k)/\Gamma(r)=(r+k-1)(r+k-2)\cdots(r)$, whence the ratio equals $\prod_{i=1}^k\frac{r+k-i}{r+\lambda}$ = $\prod_{i=1}^k\frac{1+(k-i)/r}{1+\lambda/r}$. $\endgroup$
    – whuber ♦
    Commented Oct 7, 2014 at 22:11

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