Proof of Factorial Moments and Moments of Every Order of the Poisson Distribution

Throughout, write $\pi_\mu(k)=\exp(-\mu)\,\mu^{k}/k!$ for nonnegative integers $k$, with the factorial and the conventions $0!=1$ and $\mu^{0}=1$ of the Poisson distribution definition, so that $P(K=k)=\pi_\mu(k)$ for every nonnegative integer $k$ by that definition; and write $F_p(x)=\prod_{q=0}^{p-1}(x-q)$, with the finite product notation, for real $x$ — a polynomial function of $x$. Each set $\{K=k\}=K^{-1}(\{k\})$ is an event, $\{k\}$ being a Borel set and $K$ a random variable, so $\mathcal{K}=\bigcup_{k=0}^{\infty}\{K=k\}=K^{-1}(\mathbb{N}_0)$ is an event by countable-union closure, $\mathbb{N}_0$ being the set of nonnegative integers as in the Poisson definition, and $P(\mathcal{K})=1$ by the final clause of the Poisson distribution definition (a Poisson variable lies in $\mathbb{N}_0$ with probability $1$). Note that $F_p(k)\ge0$ for every nonnegative integer $k$: for $k\le p-1$ the factor $k-k$ occurs and $F_p(k)=0$, while for $k\ge p$ every factor is positive and, by induction on $p$ using the factorial recursion $n!=n\,(n-1)!$, $F_p(k)=k!/(k-p)!$.

Step 1: series formula for nonnegative polynomial moments. Let $f$ be a polynomial function with $f(k)\ge0$ for every nonnegative integer $k$. Then $f(K)$ is a random variable: a polynomial is sequentially continuous by the algebra of limits of real sequences, so measurability of sequentially continuous functions of measurable maps applies to $K$. Moreover, writing $\mathbf{1}_{D}$ for the function equal to $1$ on a set $D$ and $0$ off $D$,

Indeed, the functions $h_n=\sum_{k=0}^{n}f(k)\,\mathbf{1}_{\{K=k\}}$ are nonnegative simple functions; grouping the indices $k\le n$ that share a common value $c$ of $f$, the standard representation of $h_n$ assigns to each value $c>0$ the event $\bigcup_{k\le n:\,f(k)=c}\{K=k\}$, whose probability is $\sum_{k\le n:\,f(k)=c}P(K=k)$ by pairwise disjointness of the $\{K=k\}$ and finite additivity of $P$, so the simple-function integral of $h_n$ equals $\sum_{k=0}^{n}f(k)\,P(K=k)$, the value $0$ contributing nothing; the $h_n$ increase pointwise to $\mathbf{1}_{\mathcal{K}}f(K)$ — at $\omega\in\mathcal{K}$ with $K(\omega)=k$ one has $h_n(\omega)=f(k)$ for every $n\ge k$, and off $\mathcal{K}$ all vanish — and monotone convergence, in its supremum form, gives the display. In particular, taking $f\equiv1$: $\sup_n\sum_{k=0}^{n}\pi_\mu(k)=\mathbb{E}[\mathbf{1}_{\mathcal{K}}]=P(\mathcal{K})=1$.

Moreover, for any polynomial $f$, if $\mathbb{E}[\mathbf{1}_{\mathcal{K}}|f(K)|]<\infty$ then $f(K)$ is integrable with $\mathbb{E}[f(K)]=\mathbb{E}[\mathbf{1}_{\mathcal{K}}f(K)]$. Indeed, writing $(\cdot)^{+}$ and $(\cdot)^{-}$ for positive and negative parts, one has the pointwise decompositions $(f(K))^{\pm}=(\mathbf{1}_{\mathcal{K}}f(K))^{\pm}+(f(K))^{\pm}\,\mathbf{1}_{\mathcal{K}^{\mathrm{c}}}$, where $\mathcal{K}^{\mathrm{c}}$ is the complement of $\mathcal{K}$, an event with $P(\mathcal{K}^{\mathrm{c}})=1-P(\mathcal{K})=0$. The last summand is dominated pointwise by the $[0,\infty]$-valued function equal to $\infty$ on $\mathcal{K}^{\mathrm{c}}$ and $0$ elsewhere — the increasing pointwise limit of the simple functions $n\,\mathbf{1}_{\mathcal{K}^{\mathrm{c}}}$, of integral $n\,P(\mathcal{K}^{\mathrm{c}})=0$ — so its integral is $0$ by monotone convergence and monotonicity. By additivity (clause 1 of the same theorem), $\mathbb{E}[(f(K))^{\pm}]=\mathbb{E}[(\mathbf{1}_{\mathcal{K}}f(K))^{\pm}]$, and these are finite, being at most $\mathbb{E}[\mathbf{1}_{\mathcal{K}}|f(K)|]<\infty$ by monotonicity, since $(\mathbf{1}_{\mathcal{K}}f(K))^{\pm}\le\mathbf{1}_{\mathcal{K}}|f(K)|$ pointwise; hence $f(K)$ is integrable and the two expectations agree.

Step 2: clause (a). Apply Step 1 with $f=F_p$. For every $n\ge p$, the terms with $k\le p-1$ vanish, and for $k\ge p$ one has $F_p(k)\,\pi_\mu(k)=\exp(-\mu)\,\mu^{k}/(k-p)!=\exp(-\mu)\,\mu^{p}\,\mu^{k-p}/(k-p)!$, using $F_p(k)=k!/(k-p)!$ and $\mu^{k}=\mu^{p}\,\mu^{k-p}$ (which for $\mu=0$ holds by the convention $\mu^{0}=1$). Hence

The right-hand partial sums are nondecreasing in $n$ with supremum $\exp(-\mu)\,\mu^{p}\,\exp(\mu)$, by the defining series of the real exponential function, and $\exp(-\mu)\,\exp(\mu)=\exp(0)=1$ by clause 1 of the basic properties of the exponential function. Hence $\mathbb{E}[\mathbf{1}_{\mathcal{K}}F_p(K)]=\mu^{p}<\infty$; since $F_p(K)\ge0$ on $\mathcal{K}$, this is also $\mathbb{E}[\mathbf{1}_{\mathcal{K}}|F_p(K)|]$, so by Step 1 the random variable $F_p(K)$ is integrable with $\mathbb{E}[F_p(K)]=\mu^{p}$. This is clause (a).

Step 3: clause (b). For every nonnegative integer $k$,

if $k\le 2p-1$ then $k^{p}\le(2p)^{p}$ and $2^{p}F_p(k)\ge0$; if $k\ge 2p$ then each factor satisfies $k-q\ge k-p\ge k/2$ for $q\in\{0,\dots,p-1\}$, so $F_p(k)\ge(k/2)^{p}$ and $k^{p}\le2^{p}F_p(k)$. Hence, for every $n$, termwise,

the final step bounding each sum by its supremum, computed in Steps 1 and 2. Taking the supremum over $n$ and using Step 1 with $f(x)=x^{p}$,

On $\mathcal{K}$ one has $K^{p}=|K|^{p}=|K^{p}|$, so $\mathbb{E}[\mathbf{1}_{\mathcal{K}}|K^{p}|]<\infty$ and, by Step 1, $K^{p}$ is integrable with $\mathbb{E}[K^{p}]=\mathbb{E}[\mathbf{1}_{\mathcal{K}}K^{p}]\le(2p)^{p}+2^{p}\,\mu^{p}$. This is clause (b).\ $\square$