use poisson distribution to show that\ne(x) = 1/λ\nvar(x) = 1/λ

Explanation:

Step1: Recall Poisson distribution PMF

The probability - mass function of a Poisson distribution with parameter $\lambda$ is $P(X = k)=\frac{e^{-\lambda}\lambda^{k}}{k!}$, for $k = 0,1,2,\cdots$.

Step2: Calculate the expected value $E(X)$

The formula for the expected value of a discrete random variable $X$ is $E(X)=\sum_{k = 0}^{\infty}kP(X = k)$. [ \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!}\ &=\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!}\ &=\lambda \end{align*} ] where we used the substitution $j=k - 1$ and the fact that $\sum_{j = 0}^{\infty}\frac{\lambda^{j}}{j!}=e^{\lambda}$.

Step3: Calculate $E(X^{2})$

The formula for $E(X^{2})=\sum_{k = 0}^{\infty}k^{2}P(X = k)$. [ \begin{align*} E(X^{2})&=\sum_{k = 0}^{\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!}\ &=\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(\lambda + 1) \end{align*} ]

Step4: Calculate the variance $Var(X)$

The formula for variance is $Var(X)=E(X^{2})-[E(X)]^{2}$. [ \begin{align*} Var(X)&=\lambda(\lambda + 1)-\lambda^{2}\ &=\lambda \end{align*} ] It seems there is a mistake in the problem statement as for a Poisson distribution $E(X)=\lambda$ and $Var(X)=\lambda$, not $\frac{1}{\lambda}$.

Answer:

The problem statement has an error. For a Poisson - distributed random variable $X$ with parameter $\lambda$, $E(X)=\lambda$ and $Var(X)=\lambda$, not $\frac{1}{\lambda}$.