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

Explanation:

Step1: Recall Poisson distribution PMF

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 P(X = k)\ &=\sum_{k = 0}^{\infty}k\cdot\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\cdot1=\lambda \end{align*} ]

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

[ \begin{align*} E(X^{2})&=\sum_{k = 0}^{\infty}k^{2}\cdot P(X = k)\ &=\sum_{k = 0}^{\infty}k^{2}\cdot\frac{e^{-\lambda}\lambda^{k}}{k!}\ &=\sum_{k = 1}^{\infty}k\cdot\frac{e^{-\lambda}\lambda^{k}}{(k - 1)!}\ &=\lambda\sum_{k = 1}^{\infty}k\cdot\frac{e^{-\lambda}\lambda^{k - 1}}{(k - 1)!}\ &=\lambda\left(\sum_{k = 1}^{\infty}(k - 1)\cdot\frac{e^{-\lambda}\lambda^{k - 1}}{(k - 1)!}+\sum_{k = 1}^{\infty}\frac{e^{-\lambda}\lambda^{k - 1}}{(k - 1)!}\right)\ &=\lambda(\lambda + 1) \end{align*} ]

Step4: Calculate the variance (Var(X))

[ \begin{align*} Var(X)&=E(X^{2})-(E(X))^{2}\ &=\lambda(\lambda + 1)-\lambda^{2}\ &=\lambda \end{align*} ]

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

Answer:

The problem statement (E(X)=\frac{1}{\lambda}) and (Var(X)=\frac{1}{\lambda}) for a Poisson - distributed random variable is incorrect. The correct results are (E(X)=\lambda) and (Var(X)=\lambda).