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

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 $k = 0,1,2,\cdots$, and the expected - value formula is $E(X)=\sum_{k = 0}^{\infty}kP(X = k)$.

Step2: Calculate the expected value

[ \begin{align*} E(X)&=\sum_{k = 0}^{\infty}k\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}e^{\lambda}=\lambda \end{align*} ]

Step3: Recall the formula for variance

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

Step4: Calculate the variance

Since $E(X)=\lambda$, then $Var(X)=E(X^{2})-[E(X)]^{2}=(\lambda^{2}+\lambda)-\lambda^{2}=\lambda$.

There seems to be a mistake in the problem statement as for a Poisson distribution $E(X)=\lambda$ and $Var(X)=\lambda$, not $\frac{1}{\lambda}$. But following the correct derivation steps for Poisson distribution:

Answer:

The expected value $E(X)=\lambda$ and the variance $Var(X)=\lambda$ for a Poisson - distributed random variable $X$ with parameter $\lambda$.