11.5 - Key Properties of a Negative Binomial Random Variable

11.5 - Key Properties of a Negative Binomial Random Variable

Theorem

Just as we did for a geometric random variable, on this page, we present and verify four properties of a negative binomial random variable.

The probability mass function:

$$f(x)=P(X=x)=\dbinom{x-1}{r-1} (1-p)^{x-r} p^r$$

for a negative binomial random variable $$X$$ is a valid p.m.f.

Proof

Before we start the "official" proof, it is helpful to take note of the sum of a negative binomial series:

$$(1-w)^{-r}=\sum\limits_{k=0}^\infty \dbinom{k+r-1}{r-1} w^k$$

Now, for the proof:

Theorem

The moment generating function of a negative binomial random variable $$X$$ is:

$$M(t)=E(e^{tX})=\dfrac{(pe^t)^r}{[1-(1-p)e^t]^r}$$

for $$(1-p)e^t<1$$.

Proof

As always, the moment generating function is defined as the expected value of $$e^{tX}$$. In the case of a negative binomial random variable, the m.g.f. is then:

$$M(t)=E(e^{tX})=\sum\limits_{x=r}^\infty e^{tx} \dbinom{x-1}{r-1} (1-p)^{x-r} p^r$$

Now, it's just a matter of massaging the summation in order to get a working formula. We start by effectively multiplying the summands by 1, and thereby not changing the overall sum:

$$M(t)=E(e^{tX})=\sum\limits_{x=r}^\infty e^{tx} \dbinom{x-1}{r-1} (1-p)^{x-r} p^r \times \dfrac{(e^t)^r}{(e^t)^r}$$

Now, since $$p^r$$ and $$(e^t)^r$$ do not depend on $$x$$, they can be pulled through the summation. And, since the $$(e^t)^r$$ that remains sits in the denominator, it can get moved into the numerator by writing is as$$(e^t)^{-r}$$:

$$M(t)=E(e^{tX})=p^r(e^t)^r \sum\limits_{x=r}^\infty e^{tx} \dbinom{x-1}{r-1} (1-p)^{x-r} (e^t)^{-r}$$

Now, the $$p^r$$ and $$(e^t)^r$$ can be pulled together as $$(pe^t)^r$$. And, $$e^{tx}$$ and $$(e^t)^r$$ can be pulled together to get $$(e^t)^{x-r}$$:

$$M(t)=E(e^{tX})=(pe^t)^r \sum\limits_{x=r}^\infty \dbinom{x-1}{r-1} (1-p)^{x-r} (e^t)^{x-r}$$

And, $$(1-p)^{x-r}$$ and $$(e^t)^{x-r}$$ can be pulled together to get $$[(1-p)e^t]^{x-r}$$:

$$M(t)=E(e^{tX})=(pe^t)^r \sum\limits_{x=r}^\infty \dbinom{x-1}{r-1} [(1-p)e^t]^{x-r}$$

Now, let $$k=x-r$$, so that $$x=k+r$$. Changing the index on the summation, we get:

$$M(t)=E(e^{tX})=(pe^t)^r \sum\limits_{k=0}^\infty \dbinom{k+r-1}{r-1}[(1-p)e^t]^k$$

Now, we should be able to recognize the summation as a negative binomial series with $$w=(1-p)e^t$$. Using what we know about the sum of a negative binomial series, the m.g.f. is then:

$$M(t)=E(e^{tX})=(pe^t)^r [1-(1-p)e^t]^{-r}$$

which can be rewritten as:

$$M(t)=E(e^{tX})=\dfrac{(pe^t)^r}{[1-(1-p)e^t]^r}$$

Now, recall that the m.g.f. exists only if it is finite. So, all we need to do is note when $$M(t)$$ is finite. Well, that happens when $$(1-p)e^t<1$$, or equivalently when $$t<-\ln (1-p)$$. And the proof is complete...whewwww!

Theorem

The mean of a negative binomial random variable $$X$$ is:

$$\mu=E(X)=\dfrac{r}{p}$$

Theorem

The variance of a negative binomial random variable $$X$$ is:

$$\sigma^2=Var(x)=\dfrac{r(1-p)}{p^2}$$

Proof

Since we used the m.g.f. to find the mean, let's use it to find the variance as well. That is, let's use:

$$\sigma^2=M''(0)-[M'(0)]^2$$

The only problem is that finding the second derivative of $$M(t)$$ is even messier than the first derivative of $$M(t)$$. Let me cheat a bit then. Let me leave it to you to verify that the second derivative of the m.g.f. of the negative binomial is:

$$M''(t)=r(pe^t)^r(-r-1)[1-(1-p)e^t]^{-r-2}[-(1-p)e^t]+r^2(pe^t)^{r-1}(pe^t)[1-(1-p)e^t]^{-r-1}$$

Now, with my shortcut taken, let's use it to evaluate the second derivative of the m.g.f. at $$t=0$$:

Now, for the final calculation:

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