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Binomial Distribution


The probability of $n$ successes in $N$ Bernoulli Trials is

P(n\vert N) = {N\choose n}p^n(1-p)^{N-n} = {N!\over n!(N-n)!} p^nq^{N-n}.
\end{displaymath} (1)

The probability of obtaining more successes than the $n$ observed is
P=\sum_{k=n+1}^N {N\choose k}p^k(1-p)^{N-k}=I_p(n+1,N-n),
\end{displaymath} (2)

I_x(a,b)\equiv {B(x; a,b)\over B(a,b)},
\end{displaymath} (3)

$B(a,b)$ is the Beta Function, and $B(x; a,b)$ is the incomplete Beta Function. The Characteristic Function is
\end{displaymath} (4)

The Moment-Generating Function $M$ for the distribution is
$\displaystyle M(t)$ $\textstyle =$ $\displaystyle \langle e^{tn}\rangle = \sum_{n=0}^N e^{tn}{N\choose n}p^nq^{N-n}$  
  $\textstyle =$ $\displaystyle \sum_{n=0}^N {N\choose n}(pe^t)^n(1-p)^{N-n}$  
  $\textstyle =$ $\displaystyle [pe^t+(1-p)]^N$ (5)
$\displaystyle M'(t)$ $\textstyle =$ $\displaystyle N[pe^t+(1-p)]^{N-1}(pe^t)$ (6)
$\displaystyle M''(t)$ $\textstyle =$ $\displaystyle N(N-1)[pe^t+(1-p)]^{N-2}(pe^t)^2$  
  $\textstyle \phantom{=}$ $\displaystyle +N[pe^t+(1-p)]^{N-1}(pe^t).$ (7)

The Mean is
\mu = M'(0) = N(p+1-p)p = Np.
\end{displaymath} (8)

The Moments about 0 are
$\displaystyle \mu_1'$ $\textstyle =$ $\displaystyle \mu=Np$ (9)
$\displaystyle \mu_2'$ $\textstyle =$ $\displaystyle Np(1-p+Np)$ (10)
$\displaystyle \mu_3'$ $\textstyle =$ $\displaystyle Np(1-3p+3Np+2p^2-3NP^2+N^2p^2)$ (11)
$\displaystyle \mu_4'$ $\textstyle =$ $\displaystyle Np(1-7p+7Np+12p^2-18Np^2+6N^2p^2$  
  $\textstyle \phantom{=}$ $\displaystyle -6p^3+11Np^3-6N^2p^3+N^3p^3),$ (12)

so the Moments about the Mean are
$\displaystyle \mu_2$ $\textstyle =$ $\displaystyle \sigma^2 = [N(N-1)p^2+Np]-(Np)^2$  
  $\textstyle =$ $\displaystyle N^2p^2-Np^2+Np-N^2p^2$  
  $\textstyle =$ $\displaystyle Np(1-p) = Npq$ (13)
$\displaystyle \mu_3$ $\textstyle =$ $\displaystyle \mu_3'-3\mu_2'\mu_1'+2(\mu_1)^3$  
  $\textstyle =$ $\displaystyle Np(1-p)(1-2p)$ (14)
$\displaystyle \mu_4$ $\textstyle =$ $\displaystyle \mu_4'-4\mu_3'\mu_1'+6\mu_2'(\mu_1')^2-3(\mu_1)^4$  
  $\textstyle =$ $\displaystyle Np(1-p)[3p^2(2-N)+3p(N-2)+1].$ (15)

The Skewness and Kurtosis are
$\displaystyle \gamma_1$ $\textstyle =$ $\displaystyle {\mu_3\over\sigma^3} = {Np(1-p)(1-2p)\over [Np(1-p)]^{3/2}}$  
  $\textstyle =$ $\displaystyle {1-2p\over\sqrt{Np(1-p)}}={q-p\over \sqrt{Npq}}$ (16)
$\displaystyle \gamma_2$ $\textstyle =$ $\displaystyle {\mu_4\over \sigma^4}-3={6p^2-6p+1\over Np(1-p)} = {1-6pq\over Npq}.$ (17)

An approximation to the Bernoulli distribution for large $N$ can be obtained by expanding about the value $\tilde n$ where $P(n)$ is a maximum, i.e., where ${dP/ dn} = 0$. Since the Logarithm function is Monotonic, we can instead choose to expand the Logarithm. Let $n\equiv \tilde n+\eta$, then
\ln[P(n)] = \ln [P(\tilde n)]+B_1\eta +{\textstyle{1\over 2}}B_2\eta^2+{\textstyle{1\over 3!}} B_3\eta^3+\ldots,
\end{displaymath} (18)

B_k\equiv \left[{d^k \ln[P(n)]\over dn^k}\right]_{n=\tilde n}.
\end{displaymath} (19)

But we are expanding about the maximum, so, by definition,
B_1 =\left[{d \ln[P(n)]\over dn}\right]_{n=\tilde n} = 0.
\end{displaymath} (20)

This also means that $B_2$ is negative, so we can write $B_2=-\vert B_2\vert$. Now, taking the Logarithm of (1) gives
\ln[P(n)] = \ln N!-\ln n!-\ln(N-n)!+n\ln p+(N-n)\ln q.
\end{displaymath} (21)

For large $n$ and $N-n$ we can use Stirling's Approximation
\ln(n!)\approx n\ln n-n,
\end{displaymath} (22)

$\displaystyle {d[\ln(n!)]\over dn}$ $\textstyle \approx$ $\displaystyle (\ln n+1)-1 = \ln n$ (23)
$\displaystyle {d[\ln(N-n)!]\over dn}$ $\textstyle \approx$ $\displaystyle {d\over dn} [(N-n)\ln(N-n)-(N-n)]$  
  $\textstyle =$ $\displaystyle \left[{-\ln(N-n)+(N-n) {-1\over N-n} +1}\right]$  
  $\textstyle =$ $\displaystyle -\ln(N-n),$ (24)

{d\ln[P(n)]\over dn} \approx -\ln n+\ln(N-n)+\ln p-\ln q.
\end{displaymath} (25)

To find $\tilde n$, set this expression to 0 and solve for $n$,
\ln\left({{N-\tilde n\over \tilde n} {p\over q}}\right)= 0
\end{displaymath} (26)

{N-\tilde n\over\tilde n} {p\over q} =1
\end{displaymath} (27)

(N-\tilde n)p=\tilde n q
\end{displaymath} (28)

\tilde n(q+p)=\tilde n=Np,
\end{displaymath} (29)

since $p+q=1$. We can now find the terms in the expansion
$\displaystyle B_2$ $\textstyle \equiv$ $\displaystyle \left[{d^2 \ln[P(n)]\over dn^2}\right]_{n=\tilde n}
= -{1\over \tilde n}-{1\over N-\tilde n}$  
  $\textstyle =$ $\displaystyle -{1\over Np}-{1\over N(1-p)} = -{1\over N} \left({{1\over p}+{1\over q}}\right)$  
  $\textstyle =$ $\displaystyle -{1\over N}\left({p+q\over pq}\right)=-{1\over Npq}
= -{1\over N(1-p)}$ (30)
$\displaystyle B_3$ $\textstyle \equiv$ $\displaystyle \left[{d^3 \ln[P(n)]\over dn^3}\right]_{n=\tilde n}
= {1\over \tilde n^2}-{1\over(N-\tilde n)^2}$  
  $\textstyle =$ $\displaystyle {1\over N^2p^2}-{1\over N^2q^2}
= {q^2-p^2\over N^2p^2q^2}$  
  $\textstyle =$ $\displaystyle {(1-2p+p^2)-p^2\over N^2p^2(1-p)^2}
= {1-2p\over N^2p^2(1-p)^2}$ (31)
$\displaystyle B_4$ $\textstyle \equiv$ $\displaystyle \left[{d^4 \ln[P(n)]\over dn^4}\right]_{n=\tilde n}
= -{2\over \tilde n^3}-{2\over(n-\tilde n)^3}$  
  $\textstyle =$ $\displaystyle -2\left({{1\over N^3p^3}+{1\over N^3q^3}}\right)={2(p^3+q^3)\over N^3p^3q^3}$  
  $\textstyle =$ $\displaystyle {2(p^2-pq+q^2)\over N^3p^3q^3}$  
  $\textstyle =$ $\displaystyle {2[p^2-p(1-p)+(1-2p+p^2)]\over N^3p^3(1-p^3)}$  
  $\textstyle =$ $\displaystyle {2(3p^2-3p+1)\over N^3p^3(1-p^3)}.$ (32)

Now, treating the distribution as continuous,
\lim_{N\to\infty} \sum_{n=0}^N P(n) \approx \int P(n)\,dn = \int_{-\infty}^\infty P(\tilde n+\eta)\,d\eta = 1.
\end{displaymath} (33)

Since each term is of order $1/N \sim 1/\sigma^2$ smaller than the previous, we can ignore terms higher than $B_2$, so
P(n)=P(\tilde n)e^{-\vert B_2\vert\eta^2/2}.
\end{displaymath} (34)

The probability must be normalized, so
\int_{-\infty}^\infty P(\tilde n)e^{-\vert B_2\vert\eta^2/2}\,d\eta = P(\tilde n) \sqrt{2\pi\over \vert B_2\vert} = 1,
\end{displaymath} (35)

$\displaystyle P(n)$ $\textstyle =$ $\displaystyle \sqrt{\vert B_2\vert\over 2\pi} e^{-\vert B_2\vert(n-\tilde n)^2/2}$  
  $\textstyle =$ $\displaystyle {1\over \sqrt{2\pi Npq}}\mathop{\rm exp}\nolimits \left[{-{(n-Np)^2\over 2Npq}}\right].$ (36)

Defining $\sigma^2\equiv 2Npq$,
P(n) = {1\over\sigma\sqrt{2\pi}}\mathop{\rm exp}\nolimits \left[{-{(n-\tilde n)^2\over 2\sigma^2}}\right],
\end{displaymath} (37)

which is a Gaussian Distribution. For $p \ll 1$, a different approximation procedure shows that the binomial distribution approaches the Poisson Distribution. The first Cumulant is
\end{displaymath} (38)

and subsequent Cumulants are given by the Recurrence Relation
\kappa_{r+1}=pq{d\kappa_r\over dp}.
\end{displaymath} (39)

Let $x$ and $y$ be independent binomial Random Variables characterized by parameters $n, p$ and $m, p$. The Conditional Probability of $x$ given that $x+y = k$ is

$P(x=i \vert x+y=k) = {P(x=i, x+y=k)\over P(x+y=k)} = {P(x=i, y=k-i)\over P(x+y=k)} = {P(x=i)P(y = k-i)\over P(x+y=k)}$
$ = {{n\choose i}p^i(1-p)^{n-i}{m\choose k-i}p^{k-i}(1-p)^{m-(k-i)} \over {n+m\choose k}p^k(1-p)^{n+m-k}} = {{n\choose i}{m\choose k-i}\over{n+m\choose k}}.\quad$ (40)
Note that this is a Hypergeometric Distribution!

See also de Moivre-Laplace Theorem, Hypergeometric Distribution, Negative Binomial Distribution


Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 531, 1987.

Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and Vetterling, W. T. ``Incomplete Beta Function, Student's Distribution, F-Distribution, Cumulative Binomial Distribution.'' §6.2 in Numerical Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. Cambridge, England: Cambridge University Press, pp. 219-223, 1992.

Spiegel, M. R. Theory and Problems of Probability and Statistics. New York: McGraw-Hill, p. 108-109, 1992.

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© 1996-9 Eric W. Weisstein