Poisson limit theorem

Let $p\_n$ be a sequence of real numbers in $[0,1]$ such that the sequence $n\; p\_n$ converges to a finite limit $\backslash lambda$. Then:

:$\backslash lim\_\{n\backslash to\; \backslash infty\}\; \{n\; \backslash choose\; k\}\; p\_n^k\; (1-p\_n)^\{n-k\}\; =\; e^\{-\backslash lambda\}\backslash frac\{\backslash lambda^k\}\{k!\}$

Assume $\backslash lambda\; >\; 0$ (the case $\backslash lambda\; =\; 0$ is easier). Then

: $$

\begin{align}

\lim\limits_{n\rightarrow\infty}{n \choose k} p_n^k (1-p_n)^{n-k}

&= \lim_{n\to\infty}\frac{n(n-1)(n-2)\dots(n-k+1)}{k!} \left(\frac{\lambda}{n}(1+o(1))\right)^k \left(1- \frac{\lambda}{n}(1+o(1))\right)^{n-k} \\

&= \lim_{n\to\infty}\frac{n^k+O\left(n^{k-1}\right)}{k!} \frac{\lambda^k}{n^k} \left(1- \frac{\lambda}{n}(1+o(1))\right)^{n} \left(1- \frac{\lambda}{n}(1+o(1))\right)^{-k}\\

&= \lim_{n\to\infty}\frac{\lambda^k}{k!} \left(1-\frac{\lambda}{n}(1+o(1))\right)^{n}.

\end{align}

Since

:$\backslash lim\_\{n\backslash to\backslash infty\}\; \backslash left(1-\backslash frac\{\backslash lambda\}\{n\}(1+o(1))\backslash right)^\{n\}\; =\; e^\{-\backslash lambda\}$

this leaves

:$\{n\; \backslash choose\; k\}p^k\; (1-p)^\{n-k\}\; \backslash simeq\; \backslash frac\{\backslash lambda^k\; e^\{-\backslash lambda\}\}\{k!\}.$

Using Stirling's approximation, it can be written:

:$$

\begin{align}

{n \choose k}p^k (1-p)^{n-k}

&= \frac{n!}{(n-k)!k!} p^k (1-p)^{n-k}

\\ &\simeq \frac{ \sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{ \sqrt{2\pi \left(n-k\right)}\left(\frac{n-k}{e}\right)^{n-k}k!} p^k (1-p)^{n-k}

\\ &= \sqrt{\frac{n}{n-k}}\frac{n^n e^{-k}}{\left(n-k\right)^{n-k}k!}p^k (1-p)^{n-k}.

\end{align}

Letting $n\; \backslash to\; \backslash infty$ and $np\; =\; \backslash lambda$:

:$$

\begin{align}

{n \choose k}p^k (1-p)^{n-k}

&\simeq \frac{n^n\,p^k (1-p)^{n-k}e^{-k}}{\left(n-k\right)^{n-k}k!}

\\&= \frac{n^n\left(\frac{\lambda}{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k}e^{-k}}{n^{n-k}\left(1-\frac{k}{n}\right)^{n-k}k!} \\&= \frac{\lambda^k \left(1-\frac{\lambda}{n}\right)^{n-k}e^{-k}}{\left(1-\frac{k}{n}\right)^{n-k}k!}

\\ &\simeq \frac{\lambda^k \left(1-\frac{\lambda}{n}\right)^{n}e^{-k}}{\left(1-\frac{k}{n}\right)^{n}k!} .

\end{align}

As $n\; \backslash to\; \backslash infty$, $\backslash left(1-\backslash frac\{x\}\{n\}\backslash right)^n\; \backslash to\; e^\{-x\}$ so:

:$\backslash begin\{align\}$

{n \choose k}p^k (1-p)^{n-k}

&\simeq \frac{\lambda^k e^{-\lambda}e^{-k}}{e^{-k}k!}

\\&= \frac{\lambda^k e^{-\lambda}}{k!}

\end{align}

It is also possible to demonstrate the theorem through the use of ordinary generating functions of the binomial distribution:

:$$

G_\operatorname{bin}(x;p,N)

\equiv \sum_{k=0}^N \left[ \binom{N}{k} p^k (1-p)^{N-k} \right] x^k

= \Big[ 1 + (x-1)p \Big]^N

by virtue of the binomial theorem. Taking the limit $N\; \backslash rightarrow\; \backslash infty$ while keeping the product $pN\backslash equiv\backslash lambda$ constant, it can be seen:

:$$

\lim_{N\rightarrow\infty} G_\operatorname{bin}(x;p,N)

= \lim_{N\rightarrow\infty} \left[ 1 + \frac{\lambda(x-1)}{N} \right]^N

= \mathrm{e}^{\lambda(x-1)}

= \sum_{k=0}^{\infty} \left[ \frac{\mathrm{e}^{-\lambda}\lambda^k}{k!} \right] x^k

which is the OGF for the Poisson distribution. (The second equality holds due to the definition of the exponential function.)

• De Moivre–Laplace theorem
• Le Cam's theorem

{{Reflist}}

Category:Articles containing proofs

Category:Theorems in probability theory