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Poisson is a limit of Binomial when n goes to infinity with np maintained

The binomial random variable has a pdf like this: \(P_X(k)=\binom{n}{k}p^k(1-p)^{n-k},\quad k=0,1,2,...,n\) Its moment-generating function is: \[\begin{align} M_X(t)=E(e^{tX})&=\sum_{k=0}^{n}e^{tk}\binom{n}{k}p^k(1-p)^{n-k}\\ &=\sum_{k=0}^{n}\binom{n}{k}(e^tp)^k(1-p)^{n-k}\\ &=(1-p+pe^t)^n \end{align}\]

Then \(M_X^{(1)}(t)=n(1-p+pe^t)^{n-1}pe^t|_{t=0}=np=E(X)\) \[\begin{align} M_X^{(2)}(t)&=n(n-1)(1-p+pe^t)^{n-2}pe^tpe^t+n(1-p+pe^t)^{n-1}pe^t|_{t=0}\\ &=n(n-1)p^2+np=E(X^2) \end{align}\]

Then \(Var(X)=E(X^2)-(E(X))^2=n(n-1)p^2+np-(np)^2=-np^2+np=np(1-p)\)

For the binomial random variable X: \(P_X(k)=\binom{n}{k}p^k(1-p)^{n-k},\quad k=0,1,2,...,n\), if \(n\to+\infty\) with \(\lambda=np\) remains constant, then \[\begin{align} \lim_{n\to+\infty}\binom{n}{k}p^k(1-p)^{n-k}&=\lim_{n\to+\infty}\frac{n!}{k!(n-k)!}(\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{n-k}\\ &=\lim_{n\to+\infty}\frac{n!}{k!(n-k)!}\lambda^k(\frac{1}{n})^k(1-\frac{\lambda}{n})^n(1-\frac{\lambda}{n})^{-k}\\ &=\frac{\lambda^k}{k!}\lim_{n\to+\infty}\frac{n!}{(n-k)!}(\frac{1}{n})^k(\frac{n}{n-\lambda})^k(1-\frac{\lambda}{n})^n\\ &=e^{-\lambda}\frac{\lambda^k}{k!}\lim_{n\to+\infty}\frac{n!}{(n-k)!}(\frac{1}{n-\lambda})^k\\ &=e^{-\lambda}\frac{\lambda^k}{k!}\lim_{n\to+\infty}\frac{n(n-1)...(n-k+1)}{(n-\lambda)(n-\lambda)...(n-\lambda)}\\ &=e^{-\lambda}\frac{\lambda^k}{k!} \end{align}\]

The moment-generating function of Poisson random variable X is: \[\begin{align} M_X(t)=E(e^{tX})&=\sum_{k=0}^{n}e^{tk}e^{-\lambda}\frac{\lambda^k}{k!}\\ &=e^{-\lambda}\sum_{k=0}^{n}\frac{(\lambda e^t)^k}{k!}\\ &=e^{-\lambda}e^{\lambda e^t}\\ &=e^{\lambda e^t-\lambda} \end{align}\]

Then, \(M_X^{(1)}(t)=\lambda e^te^{\lambda e^t-\lambda}|_{t=0}=\lambda=E(X)\) \(M_X^{(2)}(t)=\lambda (e^te^{\lambda e^t-\lambda}+e^te^t\lambda e^{\lambda e^t-\lambda})|_{t=0}=\lambda (1+\lambda)=\lambda+\lambda^2=E(X^2)\) And \(Var(X)=E(X^2)-(E(X))^2=\lambda\)