Binomial Distribution Approximated by Poisson Distribution

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Theorem

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

Then for $\lambda = n p$, $X$ can be approximated by a Poisson distribution with parameter $\lambda$:


$\ds \lim_{n \mathop \to \infty} \binom n k p^k \paren {1 - p}^{n - k} = \frac {\lambda^k} {k!} e^{-\lambda}$


Proof

Let $X$ be as described.

Let $k \in \Z_{\ge 0}$ be fixed.

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


Then:

\(\ds \lim_{n \mathop \to \infty} \binom n k p^k \paren {1 - p}^{n - k}\) \(=\) \(\ds \lim_{n \mathop \to \infty} \binom n k \paren {\frac \lambda n}^k \paren {1 - \frac \lambda n}^n \paren {1 - \frac \lambda n}^{-k}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \frac {n^k} {k!} \paren {\frac \lambda n}^k \paren {1 - \frac \lambda n}^n \paren {1 - \frac \lambda n}^{-k}\) Limit to Infinity of Binomial Coefficient over Power
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \frac 1 {k!} \lambda^k \paren {1 + \frac {-\lambda} n}^n \paren {1 - \frac \lambda n}^{-k}\) canceling $n^k$ from numerator and denominator
\(\ds \) \(=\) \(\ds \dfrac {\lambda^k} {k!} \lim_{n \mathop \to \infty} \paren {1 + \frac {-\lambda} n}^n \paren {1 - \frac \lambda n}^{-k}\) moving $\dfrac {\lambda^k} {k!}$ outside of the limit
\(\ds \) \(=\) \(\ds \dfrac {\lambda^k} {k!} \lim_{n \mathop \to \infty} \paren {1 + \frac {-\lambda} n}^n \lim_{n \mathop \to \infty} \paren {1 - \frac \lambda n}^{-k}\) Combination Theorem for Limits of Functions/Product Rule
\(\ds \) \(=\) \(\ds \dfrac {\lambda^k} {k!} \lim_{n \mathop \to \infty} \paren {1 + \frac {-\lambda} n}^n \lim_{n \mathop \to \infty} \paren {1 - \frac k n}^{-k}\) $\lambda = n p$
\(\ds \) \(=\) \(\ds \frac {\lambda^k} {k!} e^{-\lambda} \paren {1 }\) Definition of Exponential Function
\(\ds \) \(=\) \(\ds \frac {\lambda^k} {k!} e^{-\lambda}\)


Hence the result.

$\blacksquare$


Sources