# How to prove NegativeBinomial(r,p) converges to Gamma(r,1) as p->0

Let $X\sim NegBin(r,p)$ and $Y\sim Gamma(r,1)$.

How can I prove that $pX \overset{dist}\to Y$ as $p\to 0$.

Is this statement the same as $X\overset{dist}\to Gamma(r,1/p)$.

Thanks.

• Read how a binomial distribution converges to the Poisson and apply the same techniques. The Gamma distribution is the waiting time for the $r$-th arrival in a Poisson process: the negative binomial distribution is the waiting time for the occurrence of the $r$-th success of an event of probability $p$. Commented Dec 5, 2014 at 5:13

Below is my handwritten answer: (typo: pdf)

$$P_{NB}(X)= \left ( _{k}^{k+r-1}\textrm{} \right )(1-p)^{r}p^{k}$$

$$=\frac{(k+r-1)!}{k!(r-1)!}(1-p)^{\frac{1}{p}*rp}p^{k}$$

$$=\frac{(k+r-1)(k+r-2)...(k+1)}{(r-1)!}[(1-p)^{\frac{1}{p}}]^{rp}p^{k}$$

$$\approx \frac{k^{r-1}}{\Gamma (r)}e^{-rp}p^{k}$$

The tricks are:

1.You have to use (1-p)^(1/p) is converged to e when p is small.

2.Gamma function(k) = (k+1)!

3.(r+x choose r ) converages to (r+x)(r+x-1)(r+x-2)...r / r!

You will easily derive the formula with all the three above plugged in

• just updated my answer Commented Jun 28, 2016 at 19:05
• Thanks a lot. Appreciate it. Will try to write it up rigorously. Commented May 2, 2017 at 3:44

I know this question, was asked and answered years ago, but I'm going to derive $$Gamma(t;k,\lambda)$$ from a limit of $$NegBin(r;k,p)$$ in a more general way for posterity's sake. Here $$r$$ denotes the failures before obtaining $$k$$ successes.

Starting from $$NegBin(r;k,p) = {{r+k+1} \choose {k-1}}p^k(1-p)^{r}$$, we set $$p = \lambda/n$$ and $$r = tn$$. Why $$r = tn$$? Because the number of failures goes to $$\infty$$. You can think of $$n = 1/\delta t$$ ($$\delta t$$ is the time per Bernoulli trial), so that $$t$$ is linearly proportional to the number of failures (which occur with a probability approaching 1).

This yields: $${{nt+k+1} \choose {k-1}}(\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{nt}$$

Taking the limit as $$n \rightarrow \infty$$ yields:

$$\frac{t^{k-1}\lambda^k}{(k-1)!}\lambda^k e^{-\lambda t}\delta t = Gamma(t;k,\lambda)\delta t$$.