Answer to Question #329031 in Statistics and Probability for Subham

Question

Find P.D. from B.D. using Stirling’s approximation.

Answer 1

According to the Stirling's approximation:

$n!\approx\sqrt{2\pi n}(\frac{n}{e})^n$ for large n

The mean of the binomial distribution $\mu=np\Rarr p=\frac{\mu}{n}$

$P(x=k)=C_n^kp^k(1-p)^{n-k}=\\ =\frac{n!}{k!(n-k)!}(\frac{\mu}{n})^k(1-\frac{\mu}{n})^{n-k}\approx\\ \approx\frac{\sqrt{2\pi n}(n/e)^n}{\sqrt{2\pi(n-k)}((n-k)/e)^{n-k}k!}\frac{\mu^k}{n^k}\frac{(1-\mu/n)^n}{(1-\mu/n)^{k}}$

Let's use some approximations for large n:

$(1-\frac{\mu}{n})^n\approx e^{-\mu}\\ (1-\frac{\mu}{n})^k\approx1\\ \sqrt{\frac{n}{n-k}}\approx1\\ \frac{(n/e)^n}{((n-k)/e)^{n-k}}\approx\frac{e^kn^k(1-k/n)^k}{(1-k/n)^n}\approx\frac{e^{-k}n^k}{e^{-k}}=n^k$

$P(k)\approx n^k\frac{\mu^k}{n^kk!}e^{-\mu}=\frac{\mu^ke^{-\mu}}{k!}$ - Poisson's distribution.