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,

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):

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

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,

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),