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30 year-old Entertainer or Range Artist Wesley from Drumheller, really loves vehicle, property developers properties for sale in singapore singapore and horse racing. Finds inspiration by traveling to Works of Antoni Gaudí. The law of rare events or Poisson limit theorem gives a Poisson approximation to the binomial distribution, under certain conditions. The theorem was named after Siméon Denis Poisson (1781–1840).

The theorem

If

$n\to \mathrm{\infty },p\to 0$, such that $np\to \lambda$

then

$\frac{n!}{(n-k)!k!}{p}^k(1-p{)}^{n-k}\to e^{-\lambda }\frac{{\lambda }^k}{k!}.$

Example

Suppose that in an interval [0, 1000], 500 points are placed randomly. Now what is the number of points that will be placed in [0, 10]?

The probabilistically precise way of describing the number of points in the sub-interval would be to describe it as a binomial distribution $p_n(k)$.

If we look here, the probability (that a random point will be placed in the sub-interval) is $p=10/1000=0.01$. Here $n=500$ so that $np=5$.

That is, the probability that $k$ points lie in the sub-interval is

$p_n(k)=\frac{n!}{(n-k)!k!}{p}^k(1-p{)}^{n-k}.$

But using the Poisson Theorem we can approximate it as

$e^{-\lambda }\frac{{\lambda }^k}{k!}=e^{-5}\frac{5^k}{k!}.$

Proofs

Accordingly to factorial's rate of growth, we replace factorials of large numbers with approximations:

$\frac{n!}{(n-k)!k!}{p}^k(1-p{)}^{n-k}\to \frac{\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n}{\sqrt{2\pi (n-k)}{\left(\frac{n-k}{e}\right)}^{n-k}k!}{p}^k(1-p{)}^{n-k}.$

After simplifying the fraction:

$\frac{\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n}{\sqrt{2\pi (n-k)}{\left(\frac{n-k}{e}\right)}^{n-k}k!}{p}^k(1-p{)}^{n-k}\to \frac{\sqrt{n}{n}^n{p}^k(1-p{)}^{n-k}}{\sqrt{n-k}{(n-k)}^{n-k}{e}^{k}k!}\to \frac{n^n{p}^k(1-p{)}^{n-k}}{{(n-k)}^{n-k}{e}^{k}k!}.$

After using the condition $np\to \lambda$:

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

Apply, that due to $n\to \mathrm{\infty }$ we get ${(1+\frac{x}{n})}^n\to e^x$:

$\frac{{\lambda }^k{(1-\frac{\lambda }{n})}^n}{{(1-\frac{k}{n})}^n{e}^{k}k!}\to \frac{{\lambda }^k{e}^{-\lambda }}{e^{-k}{e}^{k}k!}=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$

Q.E.D.

Alternative Proof

Another proof is possible without needing approximations for the factorials. Since $np=\lambda$, we can rewrite $p=\lambda /n$. We now have:

$\underset{n\to \mathrm{\infty }}{\lim }\frac{n!}{(n-k)!k!}{\left(\frac{\lambda }{n}\right)}^k{(1-\frac{\lambda }{n})}^{n-k}=\underset{n\to \mathrm{\infty }}{\lim }\frac{n(n-1)(n-2)\dots (n-k+1)}{k!}\frac{{\lambda }^k}{n^k}{(1-\frac{\lambda }{n})}^{n-k}$

Taking each of these terms in sequence, $n(n-1)(n-2)\dots (n-k+1)=n^k+O\left(n^{k-1}\right)$, meaning that $\underset{n\to \mathrm{\infty }}{\lim }\frac{n(n-1)(n-2)\dots (n-k+1)}{n^{k}k!}=\frac{1}{k!}$.

Now ${(1-\frac{\lambda }{n})}^{n-k}={(1-\frac{\lambda }{n})}^n{(1-\frac{\lambda }{n})}^{-k}$. The first portion of this converges to $e^{-\lambda }$, and the second portion goes to 1, as $\underset{n\to \mathrm{\infty }}{\lim }{(1-\frac{\lambda }{n})}^{-k}=\underset{n\to \mathrm{\infty }}{\lim }{(1-0)}^{-k}=1$

This leaves us with $\frac{1}{k!}{\lambda }^k{e}^{-\lambda }$. Q.E.D.

Proof using Ordinary Generating Functions

It is also possible to demonstrate the theorem through the use of Ordinary Generating Functions (OGF). Indeed, the OGF of the binomial distribution is

$G_{\mathrm{b}\mathrm{i}\mathrm{n}}(x;p,N)\equiv \sum _{k=0}^N[(\genfrac{}{}{0ex}{}{N}{k})p^k(1-p{)}^{N-k}]x^k=[1+(x-1)p{]}^N$

by virtue of the Binomial Theorem. Taking the limit $N\to \mathrm{\infty }$ while keeping the product $pN\equiv \lambda$ constant, we find

$\underset{N\to \mathrm{\infty }}{\lim }{G}_{\mathrm{b}\mathrm{i}\mathrm{n}}(x;p,N)=\underset{N\to \mathrm{\infty }}{\lim }[1+\frac{\lambda (x-1)}{N}{]}^N={\mathrm{e}}^{\lambda (x-1)}=\sum _{k=0}^{\mathrm{\infty }}\left[\frac{{\mathrm{e}}^{-\lambda }{\lambda }^k}{k!}\right]x^k$

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

See also

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

References

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.