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:
| \(\ds \map \Pr {X = k}\) | \(=\) | \(\ds \binom n k p^k \paren {1 - p}^{n - k}\) | ||||||||||||
| \(\ds \) | \(\simeq\) | \(\ds \frac {n^k} {k!} \paren {\frac \lambda n}^k \paren {1 - \frac \lambda n}^n \paren {1 - \frac \lambda n}^{-k}\) | When $n >> k$, it is a reasonable approximation for $\dbinom n k$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {k!} \lambda^k \paren {1 + \frac {-\lambda} n}^n \paren {1 - \frac \lambda n}^{-k}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {k!} \lambda^k \paren {1 + \frac {-\lambda} n}^n\) | as $1 - p = \paren {1 - \dfrac \lambda n}$ is very close to $1$ | |||||||||||
| \(\ds \) | \(\simeq\) | \(\ds \frac 1 {k!} \lambda^k e^{-\lambda}\) | Definition of Exponential Function |
Hence the result.
$\blacksquare$
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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
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 2.2$: Examples