Stirling Approximation

Stirling Approximation

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Example

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

where nn is the number of heads, 2n2n is the number of trials, and pp is the probability of success.

b(n, 2n, p)=(2nn)pn(1−p)2n−nb(n, ~2n, ~p) = {2n \choose n} p^n (1-p)^{2n-n} =(2n)!n!n!pnqn= {(2n)! \over {n!n!}} p^n q^n

By Stirling's approximation,

≈2π 2n 2n2n e−2n pn qn2πn nn e−n\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)πn          ■= {(4pq) \over \sqrt{\pi n}} ~~~~~~~~~~ \blacksquare
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