The Negative Binomial Distribution: pmf, mgf, mean and variance.

Consider the situation where one performs a number Bernoulli trials, each trial has a probability of success $p$ , and
trials continue until the $r$ th successs occurs. Let $X$ be the random variable which is the number of trials up to
and including the $r$ th success. This means that the range of X is the set $\{ r, r+1, \ldots \}$ . Then the pmf would be given by ${{x-1}\choose{r-1}} p^r q^{x-r}.$

Note that there are r successes and x-r previous failures, with the last success at a fixed rth position. Thus the number of possible outcomes is the number of combinations of selecting $(x-1)$ objects taken $(r-1)$ at a time.

The moment generating function of the negative binomial distribution

What is the mgf of the negative binomial distribution? Let us compute the value of

$M_Xt = E(tX) = \sum _{x=r}^{\infty} e^{tx}{{x-1}\choose{r-1}} p^r q^{x-r}.$

Writing the first few terms of the expansion, we get

$M_Xt = e^{tr}{{r-1}\choose{r-1}} p^r q^{r-r} +e^{t(r+1)}{{r}\choose{r-1}} p^r q^{r+1-r} +e^{t(r+2)}{{r+1}\choose{r-1}} p^r q^{r+2-r}+\ldots$
which simplifies to

$E(tX) = e^{tr} p^r + e^{t(r+1)} {r \choose {1}} p^r q +e^{t(r+2)} {(r+1) \choose{2}} p^r q^{2}+\ldots$

and which upon factoring out $p^re^{tr} = (pe^t)^r$ and further simplication results in

$(pe^t)^r [ 1 + {r \choose {1}} (q e^t) + {(r+1) \choose{2}} (q e^t)^{2}+\ldots]$
It will be shown later that the bracketed terms is equivalent to
$(1-qe^t)^{-r}$ , see the blog entry A negative binomial series identity and therefore the moment generating function of the negative binomial distribution
is given by
$M_Xt = \frac{(pe^t)^r}{(1-qe^t)^r}$

Note that the numerator in the formula of 
Spiegel's Statistics in page. 118 should be raised to the power
.

The mean of the Negative Binomial Distribution

The expectation or mean of the negative binomial distribution with the pmf and mgf above is obtained by differentiating the mgf wrt t and setting t to zero: $E(X) = \mu= M'_X(t)|{t = 0}.$

When differentiated, the derivative is
$\frac{(1-qe^t)^r r(pe^t)^{r-1}pe^t + (pe^t)^r r qe^t{(1-qe^t)^{r-1}}}{(1-qe^t)^{2r}}$
and the value at t=0 is

$\frac{r p^{2r-1}( p+q)}{p^2r}$ which collapses to $\mu = \frac{r}{p}$

The variance of the Negative Binomial Distribution

The second moment or $E(X^2)$ of the negative binomial distribution with the pmf and mgf above is obtained by differentiating the mgf twice wrt t and setting t to zero and the variance is computed as

$V(X) = E(X^2) -[E(X)]^2.$

We will leave as an exercise (at the moment since it is so tedious) that the variance is given by

$V(X) = \frac{rq}{p^2}$ .

This entry is subject to review but the final formulas are all right.

Revisions.

Feb. 20, 2010: We missed the square in the denominator! We will redo the presentation for the variance.

This entry was posted on Friday, January 8th, 2010 at 6:41 pm and is filed under Applied Mathematics, Statistics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

5 Responses to “The Negative Binomial Distribution: pmf, mgf, mean and variance.”

Ramadan Ally Says:
January 21st, 2011 at 4:41 pm
It will be better if you will provide to us the derivation of variance of negative binomial distribution by using raw moment.. (I'm from Tanzania}
ernie Says:
January 21st, 2011 at 4:49 pm
Thanks, will take a look at this again before January is over. Reader feedback is important to us.
Lee Corbin Says:
April 23rd, 2011 at 4:40 am
Very nice article. There is a typo in the
equation following "which simplifies to"
in the exponent of the 3rd term. Thank you.
DAV Says:
January 1st, 2013 at 8:13 am
This is a very big effort. It has simplified my academic work. Keep it up
kenny Says:
April 28th, 2013 at 6:36 pm
thanks.

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The Negative Binomial Distribution: pmf, mgf, mean and variance.

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