use poisson distribution to show that
e(x) = 1/x
var(x) = 1/x

Question

Accepted Answer

# Explanation:
## Step1: Recall the formula for the Poisson distribution
The probability mass function of a Poisson random variable $X$ with parameter $\lambda$ is $P(X = k)=\frac{e^{-\lambda}\lambda^{k}}{k!}$, for $k = 0,1,2,\cdots$

## Step2: Calculate the expected value $E(X)$
$$
\begin{align*}
E(X)&=\sum_{k = 0}^{\infty}k\cdot\frac{e^{-\lambda}\lambda^{k}}{k!}\
&=\lambda\sum_{k = 1}^{\infty}\frac{e^{-\lambda}\lambda^{k - 1}}{(k - 1)!}\
\end{align*}
$$
Let $m=k - 1$, then $\sum_{k = 1}^{\infty}\frac{e^{-\lambda}\lambda^{k - 1}}{(k - 1)!}=\sum_{m=0}^{\infty}\frac{e^{-\lambda}\lambda^{m}}{m!}=1$ (since it is the sum of a Poisson probability mass function over all possible values). So $E(X)=\lambda$

## Step3: Calculate $E(X^{2})$
$$
\begin{align*}
E(X^{2})&=\sum_{k = 0}^{\infty}k^{2}\cdot\frac{e^{-\lambda}\lambda^{k}}{k!}\
&=\sum_{k = 0}^{\infty}k(k - 1)\cdot\frac{e^{-\lambda}\lambda^{k}}{k!}+\sum_{k = 0}^{\infty}k\cdot\frac{e^{-\lambda}\lambda^{k}}{k!}\
&=\lambda^{2}\sum_{k = 2}^{\infty}\frac{e^{-\lambda}\lambda^{k - 2}}{(k - 2)!}+\lambda\
\end{align*}
$$
Let $n=k - 2$, then $\sum_{k = 2}^{\infty}\frac{e^{-\lambda}\lambda^{k - 2}}{(k - 2)!}=\sum_{n=0}^{\infty}\frac{e^{-\lambda}\lambda^{n}}{n!}=1$
So $E(X^{2})=\lambda^{2}+\lambda$

## Step4: Calculate the variance $Var(X)$
$$
\begin{align*}
Var(X)&=E(X^{2})-(E(X))^{2}\
&=\lambda^{2}+\lambda-\lambda^{2}\
&=\lambda
\end{align*}
$$

# Answer:
The expected value $E(X)=\lambda$ and the variance $Var(X)=\lambda$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the formula for the Poisson distribution

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

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

Step4: Calculate the variance \(Var(X)\)

Answer: