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Negative Binomial Distribution

Also known as the Pascal Distribution and Pólya Distribution. The probability of $r-1$ successes and $x$ failures in $x+r-1$ trials, and success on the $(x+r)$th trial is


\begin{displaymath}
p\left[{{x+r-1\choose r-1} p^{r-1}(1-p)^{[(x+r-1)-(r-1)]}}\r...
...se r-1}p^{r-1}(1-p)^x}\right]p = {x+r-1\choose r-1}p^r(1-p)^x,
\end{displaymath} (1)

where ${n\choose k}$ is a Binomial Coefficient. Let
$\displaystyle P$ $\textstyle =$ $\displaystyle {1-p\over p}$ (2)
$\displaystyle Q$ $\textstyle =$ $\displaystyle {1\over p}.$ (3)

The Characteristic Function is given by
\begin{displaymath}
\phi(t)=(Q-Pe^{it})^{-r},
\end{displaymath} (4)

and the Moment-Generating Function by
\begin{displaymath}
M(t) = \langle e^{tx}\rangle = \sum_{x=0}^\infty e^{tx}{x+r-1\choose r-1}p^r(1-p)^x,
\end{displaymath} (5)

but, since ${N\choose n}={N\choose N-m}$,


$\displaystyle M(t)$ $\textstyle =$ $\displaystyle p^r \sum_{x=0}^\infty{x+r-1\choose x}[(1-p)e^t]^x$  
  $\textstyle =$ $\displaystyle p^r[1-(1-p)e^t]^{-r}$ (6)
$\displaystyle M'(t)$ $\textstyle =$ $\displaystyle p^{r}(-r)[1-(1-p)e^t]^{-r-1}(p-1)e^t$  
  $\textstyle =$ $\displaystyle p^{r}(1-p)r[1-(1-p)e^t]^{-r-1}e^t$ (7)
$\displaystyle M''(t)$ $\textstyle =$ $\displaystyle (1-p)r p^r(1-e^t+p e^t)^{-r-2}(-1-e^t r+e^t pr)e^t$ (8)
$\displaystyle M'''(t)$ $\textstyle =$ $\displaystyle (1-p)r p^r(1-e^t+e^t p)^{-r-3}[1+e^t(1-p+3r-3pr)+r^2 e^{2t}(1-p)^2]e^t.$ (9)

The Moments about zero $\mu_n'=M^{n}(0)$ are therefore
$\displaystyle \mu_1'$ $\textstyle =$ $\displaystyle \mu={r(1-p)\over p}={rq\over p}$ (10)
$\displaystyle \mu_2'$ $\textstyle =$ $\displaystyle {r(1-p)[1-r(p-1)]\over p^2}={rq(1-rq)\over p^2}$ (11)
$\displaystyle \mu_3'$ $\textstyle =$ $\displaystyle {(1-p)r(2-p+3r-3pr+r^2-2pr^2+p^2r^2\over p^3}$ (12)
$\displaystyle \mu_4'$ $\textstyle =$ $\displaystyle {(-1 + p) r (-6 + 6 p - p^2 - 11 r + 15 p r - 4 p^2r - 6 r^2\over p^4}$  
  $\textstyle \phantom{=}$ $\displaystyle + {12 p r^2 - 6 p^2 r^2 - r^3 + 3 p r^3 - 3 p^2 r^3 + p^3 r^3)\over p^4}.$ (13)

(Beyer 1987, p. 487, apparently gives the Mean incorrectly.) The Moments about the mean are
$\displaystyle \mu_2$ $\textstyle =$ $\displaystyle \sigma^2 = {r(1-p)\over p^2}$ (14)
$\displaystyle \mu_3$ $\textstyle =$ $\displaystyle {r(2-3p+p^2)\over p^3}={r(p-1)(p-2)\over p^3}$ (15)
$\displaystyle \mu_4$ $\textstyle =$ $\displaystyle {r(1-p)(6-6p+p^2+3r-3pr)\over p^4}.$ (16)

The Mean, Variance, Skewness and Kurtosis are then
$\displaystyle \mu$ $\textstyle =$ $\displaystyle {r(1-p)\over p}$ (17)
$\displaystyle \gamma_1$ $\textstyle =$ $\displaystyle {\mu_3\over\sigma^3}={r(p-1)(p-2)\over p^3} \left[{p^2\over r(1-p)}\right]^{3/2}$  
  $\textstyle =$ $\displaystyle {r(2-p)(1-p)\over p^3} {p^3\over r(1-p)\sqrt{1-p}}$  
  $\textstyle =$ $\displaystyle {2-p\over\sqrt{r(1-p)}}$ (18)
$\displaystyle \gamma_2$ $\textstyle =$ $\displaystyle {\mu_4\over\sigma^4}-3$  
  $\textstyle =$ $\displaystyle {-6 + 6p - p^2 - 3 r + 3p r\over (p-1) r},$ (19)

which can also be written
$\displaystyle \mu$ $\textstyle =$ $\displaystyle nP$ (20)
$\displaystyle \mu_2$ $\textstyle =$ $\displaystyle nPQ$ (21)
$\displaystyle \gamma_1$ $\textstyle =$ $\displaystyle {Q+P\over\sqrt{rPQ}}$ (22)
$\displaystyle \gamma_2$ $\textstyle =$ $\displaystyle {1+6PQ\over rPQ}-3.$ (23)

The first Cumulant is
\begin{displaymath}
\kappa_1=nP,
\end{displaymath} (24)

and subsequent Cumulants are given by the recurrence relation
\begin{displaymath}
\kappa_{r+1}=PQ{d\kappa_r\over dQ}.
\end{displaymath} (25)


References

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 533, 1987.

Spiegel, M. R. Theory and Problems of Probability and Statistics. New York: McGraw-Hill, p. 118, 1992.



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© 1996-9 Eric W. Weisstein
1999-05-25