1. **Stating the problem:** We are given the probability mass function (PMF) of a Poisson random variable $X$: $$P(X = x) = \frac{e^{-\mu} \mu^x}{x!}$$ where $\mu$ is the parameter. We need to show: (i) $E(X) = \mu$ (ii) $\mathrm{Var}(X) = \mu$ 2. **Recall formulas and important rules:** - The expectation (mean) of a discrete random variable $X$ is: $$E(X) = \sum_{x=0}^\infty x P(X=x)$$ - The variance is: $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2$$ - For the Poisson distribution, the PMF is as given. 3. **Calculate $E(X)$:** $$E(X) = \sum_{x=0}^\infty x \frac{e^{-\mu} \mu^x}{x!}$$ Note that when $x=0$, the term is zero because of the factor $x$. Rewrite the sum starting from $x=1$: $$E(X) = e^{-\mu} \sum_{x=1}^\infty x \frac{\mu^x}{x!}$$ Simplify $x \frac{\mu^x}{x!} = \mu \frac{\mu^{x-1}}{(x-1)!}$: $$E(X) = e^{-\mu} \mu \sum_{x=1}^\infty \frac{\mu^{x-1}}{(x-1)!}$$ Change index $k = x-1$: $$E(X) = e^{-\mu} \mu \sum_{k=0}^\infty \frac{\mu^k}{k!}$$ Recognize the series as the expansion of $e^{\mu}$: $$\sum_{k=0}^\infty \frac{\mu^k}{k!} = e^{\mu}$$ Therefore: $$E(X) = e^{-\mu} \mu e^{\mu} = \mu$$ 4. **Calculate $E(X^2)$:** We use the identity: $$E(X(X-1)) = \sum_{x=0}^\infty x(x-1) P(X=x)$$ Calculate: $$E(X(X-1)) = e^{-\mu} \sum_{x=0}^\infty x(x-1) \frac{\mu^x}{x!}$$ For $x=0$ and $x=1$, terms are zero. Start from $x=2$: $$E(X(X-1)) = e^{-\mu} \sum_{x=2}^\infty x(x-1) \frac{\mu^x}{x!}$$ Rewrite $x(x-1) \frac{\mu^x}{x!} = \mu^2 \frac{\mu^{x-2}}{(x-2)!}$: $$E(X(X-1)) = e^{-\mu} \mu^2 \sum_{x=2}^\infty \frac{\mu^{x-2}}{(x-2)!}$$ Change index $k = x-2$: $$E(X(X-1)) = e^{-\mu} \mu^2 \sum_{k=0}^\infty \frac{\mu^k}{k!} = e^{-\mu} \mu^2 e^{\mu} = \mu^2$$ 5. **Use $E(X^2) = E(X(X-1)) + E(X)$:** $$E(X^2) = \mu^2 + \mu$$ 6. **Calculate variance:** $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = (\mu^2 + \mu) - \mu^2 = \mu$$ **Final answers:** (i) $E(X) = \mu$ (ii) $\mathrm{Var}(X) = \mu$