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

Explanation:

Step1: Recall Poisson distribution formula

The probability mass function of a Poisson - distributed random variable (X) with parameter (\lambda) is (P(X = k)=\frac{e^{-\lambda}\lambda^{k}}{k!}), (k = 0,1,2,\cdots). The expected value (E(X)) is calculated as (E(X)=\sum_{k = 0}^{\infty}k\cdot P(X = k)). [ \begin{align*} E(X)&=\sum_{k = 0}^{\infty}k\cdot\frac{e^{-\lambda}\lambda^{k}}{k!}\ &=e^{-\lambda}\sum_{k = 1}^{\infty}\frac{\lambda^{k}}{(k - 1)!}\ &=\lambda e^{-\lambda}\sum_{j = 0}^{\infty}\frac{\lambda^{j}}{j!}\quad(\text{let }j=k - 1)\ &=\lambda e^{-\lambda}\cdot e^{\lambda}=\lambda \end{align*} ]

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}\cdot\frac{e^{-\lambda}\lambda^{k}}{k!}\ &=e^{-\lambda}\sum_{k = 1}^{\infty}k\cdot\frac{\lambda^{k}}{(k - 1)!}\ &=\lambda e^{-\lambda}\sum_{k = 1}^{\infty}k\cdot\frac{\lambda^{k - 1}}{(k - 1)!}\ &=\lambda e^{-\lambda}\left(\sum_{k = 1}^{\infty}(k - 1)\frac{\lambda^{k - 1}}{(k - 1)!}+\sum_{k = 1}^{\infty}\frac{\lambda^{k - 1}}{(k - 1)!}\right)\ &=\lambda e^{-\lambda}\left(\lambda\sum_{j = 0}^{\infty}\frac{\lambda^{j}}{j!}+\sum_{j = 0}^{\infty}\frac{\lambda^{j}}{j!}\right)\quad(\text{let }j = k - 1)\ &=\lambda(\lambda + 1) \end{align*} ] Then (Var(X)=E(X^{2})-[E(X)]^{2}=\lambda(\lambda + 1)-\lambda^{2}=\lambda)

It seems there is a mistake in the problem statement as for a Poisson - distributed random variable (X) 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 - distributed random variable (X) with parameter (\lambda) are (E(X)=\lambda) and (Var(X)=\lambda), not (E(X)=\frac{1}{\lambda}) and (Var(X)=\frac{1}{\lambda}) as stated in the problem.