Binomial Distribution Approximated by Poisson Distribution

Theorem

Let $X$ be a discrete random variable which has the binomial distribution with parameters $n$ and $p$.

Suppose $n$ is "very large" and $p$ is "very small", but $np$ of a "reasonable size".

Then $X$ can be approximated by a Poisson distribution with parameter $\lambda$ where $\lambda = np$.

Proof

Let $X$ be as described.

Let $k \ge 0$ be fixed.

We write $p = \dfrac \lambda n$ and suppose that $n$ is large.

Then:

	\(\displaystyle \)	\(\displaystyle \Pr \left({X = k}\right)\)	\(=\)	\(\displaystyle \binom n k p^k \left({1 - p}\right)^{n-k}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\simeq\)	\(\displaystyle \frac {n^k} {k!} \left({\frac \lambda n}\right)^k \left({1 - \frac \lambda n}\right)^n \left({1 - \frac \lambda n}\right)^{-k}\)	\(\displaystyle \)	When $n >> k$ it's a reasonable approximation for $\displaystyle \binom n k$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {k!} \lambda^k \left({1 + \frac {-\lambda} n}\right)^n \left({1 - \frac \lambda n}\right)^{-k}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {k!} \lambda^k \left({1 + \frac {-\lambda} n}\right)^n\)	\(\displaystyle \)	as $1 - p = \left({1 - \dfrac \lambda n}\right)$ is very close to $1$
	\(\displaystyle \)	\(\displaystyle \)	\(\simeq\)	\(\displaystyle \frac 1 {k!} \lambda^k e^{-\lambda}\)	\(\displaystyle \)	Exponential as the Limit of a Sequence

Hence the result.

$\blacksquare$

Comment

Okay wise guy, exactly what constitutes "very large", "very small", and "of a reasonable size"?

Well, if $n = 10^6$ and $p = 10^{-5}$, we have $np = 10$.

That's the sort of order of magnitude we're talking about here.

Sources

• Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction (1986): $\S 2.2$: Example $11$