Derivation

The probability of \(k\) successes for binomial distribution with parameters \((n,p)\) is equal to \[ {n\choose k}p^k(1-p)^{n-k}. \] Let us find its limit when \(n\to \infty.\) \[ \lim_{n\to\infty}{n\choose k}p^k(1-p)^k=\frac{1}{k!}\lim_{n\to\infty}\frac{n!}{(n-k)!}p^k(1-p)^{n-k}= \] by Stirling formula, \(n!\sim \sqrt{2\pi n}\frac{n^n}{e^n}\) \[ =\frac{1}{k!}\lim_{n\to\infty}\frac{\sqrt{2\pi n}n^ne^{n-k}}{\sqrt{2\pi(n-k)}(n-k)^{n-k}e^n}p^k(1-p)^{n-k}= \] \[ =\frac{1}{e^kk!}\lim_{n\to\infty}\bigg(\frac{n}{n-k}\bigg)^{n-k}(np)^k(1-p)^{n-k}= \] \[ =\frac{\lambda^k}{e^kk!}\lim_{n\to\infty}\bigg(1-\frac{k}{n}\bigg)^{-(n-k)}\bigg(1-\frac{np}{n}\bigg)^{n-k}= \] use that \((1+x/n)^n\to e^x\) \[ =\frac{e^ke^{-\lambda}\lambda^k}{e^kk!}=e^{-\lambda} \frac{\lambda^k}{k!} \]

Moment generating function: \[ M(t)=e^{\lambda(e^t-1)} \]

Proof

\[ M(t)=Ee^{tX}=\sum^\infty_{k=0} e^{tk}e^{-\lambda} \frac{\lambda^k}{k!}= \] \[ =e^{-\lambda}\sum^\infty_{k=0}\frac{(\lambda e^t)^k}{k!}=e^{-\lambda}e^{\lambda e^t}=e^{\lambda(e^t-1)} \]

Expectation: \(EX=\lambda\)

Variance: \(V(X)=\lambda\)

Derivation

\[ M'(t)=\lambda e^t e^{\lambda(e^t-1)}, \ EX=M'(0)=\lambda \] \[ M''(t)=\lambda e^t e^{\lambda(e^t-1)}+\lambda^2 e^{2t} e^{\lambda(e^t-1)} \] \[ EX^2=M''(0)=\lambda+\lambda^2 \] Variance is \[ V(X)=\lambda \]