Stirling Approximation

Warning

Over a year old. May be outdated

Example

b(n, ~2n, ~p) \approx {(4pq) \over \sqrt{\pi n}}

where $n$ is the number of heads, $2n$ is the number of trials, and $p$ is the probability of success.

b(n, ~2n, ~p) = {2n \choose n} p^n (1-p)^{2n-n}

= {(2n)! \over {n!n!}} p^n q^n

By Stirling's approximation,

\approx {{\sqrt{2\pi ~ 2n} ~ {2n}^{2n} ~ e^{-2n} ~ p^n ~ q^n} \over {\sqrt{2 \pi n} ~ n^n ~ e^{-n}}}

Cleaning up,

= {(4pq) \over \sqrt{\pi n}} ~~~~~~~~~~ \blacksquare

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