Proof of Moments of the Poisson Distribution

Write $\pi_{\mu}(n)=\exp(-\mu)\mu^{n}/n!$ for $n\in\mathbb{N}_0=\mathbb{N}\cup\{0\}$, with the conventions of Poisson Distribution. As in Step 1 of the proof of Thinning: Cell Counts of a Poisson Number of Independent Points, let $\widetilde K=K$ on the event $\{K\in\mathbb{N}_0\}$ and $\widetilde K=0$ elsewhere; then $\widetilde K$ is an $\mathbb{N}_0$-valued random variable, $P(\widetilde K=K)=1$, and $P(\widetilde K=n)=\pi_{\mu}(n)$ for every $n\in\mathbb{N}_0$. Expectations are handled with Linearity and Monotonicity of the Lebesgue Integral, and we use the null-support principle recorded in the proof of Mean-Square Completeness of Square-Integrable Random Variables (Riesz-Fischer): a nonnegative random variable vanishing off an event of probability $0$ has expectation $0$.

Step 1 (moments of $\widetilde K$). For $m\in\mathbb{N}$ define the nonnegative simple functions $s_m=\sum_{n=0}^{m}n^{2}\,\mathbf{1}_{\{\widetilde K=n\}}$, where $\mathbf{1}_{E}$ is $1$ on $E$ and $0$ off $E$. The sequence $(s_m)$ is nondecreasing and converges pointwise to $\widetilde K^{2}$: at each $\omega$, once $m\ge\widetilde K(\omega)$ we have $s_m(\omega)=\widetilde K(\omega)^{2}$. By the monotone convergence theorem and the integral of simple functions,

$$\mathbb{E}[\widetilde K^{2}]=\lim_{m\to\infty}\mathbb{E}[s_m]=\lim_{m\to\infty}\sum_{n=0}^{m}n^{2}\,\pi_{\mu}(n)=\sum_{n=0}^{\infty}n^{2}\,\pi_{\mu}(n).$$

To evaluate the series, use $n^{2}=n(n-1)+n$ and split (all terms nonnegative, so the two partial-sum sequences converge separately and add, by the algebra of limits):

$$\sum_{n=0}^{\infty}\frac{n(n-1)\,\mu^{n}}{n!}=\mu^{2}\sum_{n=2}^{\infty}\frac{\mu^{n-2}}{(n-2)!}=\mu^{2}\exp(\mu),\qquad\sum_{n=0}^{\infty}\frac{n\,\mu^{n}}{n!}=\mu\sum_{n=1}^{\infty}\frac{\mu^{n-1}}{(n-1)!}=\mu\exp(\mu),$$

using the factorial recursions $n!=n(n-1)!$ and $n!=n(n-1)(n-2)!$ (for $n\ge1$, resp. $n\ge2$; the terms $n=0,1$ of the first series and $n=0$ of the second vanish), the index shifts being equalities of partial sums, and the defining series of the exponential function. Hence

$$\mathbb{E}[\widetilde K^{2}]=\exp(-\mu)\bigl(\mu^{2}+\mu\bigr)\exp(\mu)=\mu+\mu^{2}<\infty,$$

by Basic Properties of the Exponential Function. The same argument with $n$ in place of $n^{2}$ gives $\mathbb{E}[\widetilde K]=\mu$.

Step 2 (transfer to $K$). Let $G=\{K=\widetilde K\}$, so $P(\Omega\setminus G)=0$. The nonnegative random variables $K^{2}\mathbf{1}_{\Omega\setminus G}$ and $\widetilde K^{2}\mathbf{1}_{\Omega\setminus G}$ have expectation $0$ by the null-support principle, and $K^{2}\mathbf{1}_{G}=\widetilde K^{2}\mathbf{1}_{G}$ pointwise. Adding the pieces with additivity of expectation on nonnegative random variables,

$$\mathbb{E}[K^{2}]=\mathbb{E}[K^{2}\mathbf{1}_{G}]+0=\mathbb{E}[\widetilde K^{2}\mathbf{1}_{G}]=\mathbb{E}[\widetilde K^{2}]-\mathbb{E}[\widetilde K^{2}\mathbf{1}_{\Omega\setminus G}]=\mu+\mu^{2}.$$

In particular $K$ is square-integrable, hence integrable by Square-Integrable Random Variables and the Mean-Square Inner Product. For the first moment, decompose the integrable variable $K-\widetilde K$: it vanishes on $G$, so its positive and negative parts are nonnegative variables vanishing off the null event $\Omega\setminus G$, each with expectation $0$; hence $\mathbb{E}[K-\widetilde K]=0$ and $\mathbb{E}[K]=\mathbb{E}[\widetilde K]=\mu$ by linearity.

Step 3 (variance). By Expectation, Variance, and Moments, $\operatorname{Var}(K)=\mathbb{E}\bigl[(K-\mathbb{E}[K])^{2}\bigr]$. Expanding $(K-\mu)^{2}=K^{2}-2\mu K+\mu^{2}$ pointwise and using linearity (all terms integrable),

$$\operatorname{Var}(K)=\mathbb{E}[K^{2}]-2\mu\,\mathbb{E}[K]+\mu^{2}=(\mu+\mu^{2})-2\mu^{2}+\mu^{2}=\mu.\qquad\blacksquare$$