I saw a cool trick the other day involving the Poisson Distribution and Stirling's Approximation. Given a Poisson Distribution $ N $ ~ $ Poi( \lambda ) $ The probability that $N$ is equal to a given $n$ is defined to be: $$P ( N = n) = \frac { {\lambda}^n e^{n} } {n! } $$ What is the probability that $N$ is equal to it's mean? In this case, let's use $n$ as the mean of the distribution for reasons that will become clear later. Plugging $n$ into the definition of the Poisson distribution gives: $$P ( N = n) = \frac { n^n e^{n} } {n! } $$ At this point, we use Sterling's approximation. Which states that for large $n$: $n!$ ~ $ {\left( \frac { n } {e } \right) }^n \frac { 1 } { \sqrt{ 2 \pi n } }$ Plugging this into the definition of the Poission Distribution gives: $$P ( N = n) = \frac { n^n e^{n} } {{\left( \frac { n } {e } \right) }^n} \frac { 1 } { \sqrt{ 2 \pi n } } $$ Which simplifies to: $$P ( N = n) = \frac { 1 } { \sqrt{ 2 \pi n } } $$ So for large $n$ we end up with a nice result for the Probability that a Poisson Distribution will end up being equal to it's Expected Value. Convergence The convergence of the series is actually really quick. I checked the convergence for n between 1 and 50, and even by n=5, the approximation is very close, when I graphed it, the lines become indistinguishable very quickly. 
AuthorI work as a pricing actuary at a reinsurer in London. Categories
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May 2020

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