use poisson distribution to show that $e(x)=\frac{1}{lambda}$ and $var(x)=\frac{1}{lambda}$

Explanation:

Step1: Recall Poisson distribution formula

The probability mass function of a Poisson - distribution is (P(X = k)=\frac{e^{-\lambda}\lambda^{k}}{k!}), where (k = 0,1,2,\cdots) and (\lambda>0). The expected value (E(X)) is defined as (E(X)=\sum_{k = 0}^{\infty}kP(X = k)). [ \begin{align*} E(X)&=\sum_{k = 0}^{\infty}k\frac{e^{-\lambda}\lambda^{k}}{k!}\ &=\sum_{k = 1}^{\infty}k\frac{e^{-\lambda}\lambda^{k}}{k!}\quad(\text{since when }k = 0,k\frac{e^{-\lambda}\lambda^{k}}{k!}=0)\ &=\sum_{k = 1}^{\infty}\frac{e^{-\lambda}\lambda^{k}}{(k - 1)!}\ &=\lambda\sum_{k = 1}^{\infty}\frac{e^{-\lambda}\lambda^{k-1}}{(k - 1)!}\ &=\lambda e^{-\lambda}\sum_{j = 0}^{\infty}\frac{\lambda^{j}}{j!}\quad(\text{let }j=k - 1) \end{align*} ] Since (\sum_{j = 0}^{\infty}\frac{\lambda^{j}}{j!}=e^{\lambda}), then (E(X)=\lambda).

Step2: Recall the formula for variance

The variance (Var(X)=E(X^{2})-[E(X)]^{2}). First, find (E(X^{2})): [ \begin{align*} E(X^{2})&=\sum_{k = 0}^{\infty}k^{2}\frac{e^{-\lambda}\lambda^{k}}{k!}\ &=\sum_{k = 1}^{\infty}k^{2}\frac{e^{-\lambda}\lambda^{k}}{k!}\ &=\sum_{k = 1}^{\infty}k\frac{e^{-\lambda}\lambda^{k}}{(k - 1)!}\ &=\lambda\sum_{k = 1}^{\infty}k\frac{e^{-\lambda}\lambda^{k - 1}}{(k - 1)!}\ &=\lambda\sum_{j = 0}^{\infty}(j + 1)\frac{e^{-\lambda}\lambda^{j}}{j!}\quad(\text{let }j=k - 1)\ &=\lambda\left(\sum_{j = 0}^{\infty}j\frac{e^{-\lambda}\lambda^{j}}{j!}+\sum_{j = 0}^{\infty}\frac{e^{-\lambda}\lambda^{j}}{j!}\right)\ &=\lambda\left(\lambda\sum_{j = 1}^{\infty}\frac{e^{-\lambda}\lambda^{j-1}}{(j - 1)!}+1\right)\ &=\lambda(\lambda + 1) \end{align*} ] Since (E(X)=\lambda), then (Var(X)=E(X^{2})-[E(X)]^{2}=\lambda(\lambda + 1)-\lambda^{2}=\lambda).

There is a mistake in the problem statement. For a Poisson - distribution with parameter (\lambda), (E(X)=\lambda) and (Var(X)=\lambda), not (E(X)=\frac{1}{\lambda}) and (Var(X)=\frac{1}{\lambda}).

Answer:

The expected value (E(X)) and variance (Var(X)) of a Poisson - distribution with parameter (\lambda) are (E(X)=\lambda) and (Var(X)=\lambda), not (\frac{1}{\lambda}) as stated in the problem.