Question #329031

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



1
Expert's answer
2022-04-17T23:23:51-0400

According to the Stirling's approximation:

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


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


P(x=k)=Cnkpk(1p)nk==n!k!(nk)!(μn)k(1μn)nk2πn(n/e)n2π(nk)((nk)/e)nkk!μknk(1μ/n)n(1μ/n)kP(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:

(1μn)neμ(1μn)k1nnk1(n/e)n((nk)/e)nkeknk(1k/n)k(1k/n)neknkek=nk(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)nkμknkk!eμ=μkeμk!P(k)\approx n^k\frac{\mu^k}{n^kk!}e^{-\mu}=\frac{\mu^ke^{-\mu}}{k!} - Poisson's distribution.


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