Poisson Distribution

Let $\mathbb{N}$ be the set of natural numbers and write $\mathbb{N}_0=\mathbb{N}\cup\{0\}$ for the set of nonnegative integers; let $\mathbb{R}$ be the set of real numbers and $\mathcal{B}(\mathbb{R})$ the Borel $\sigma$-algebra. We use the factorial $k!$ for $k\in\mathbb{N}$, extended by the convention $0!=1$ (the cited definition covers only $k\ge1$), together with the convention $\mu^{0}=1$.

Fix a real number $\mu\ge0$. The Poisson distribution with parameter $\mu$ is the function

$$P_\mu:\mathcal{B}(\mathbb{R})\to[0,1],\qquad P_\mu(B)=\exp(-\mu)\sum_{k\in\mathbb{N}_0,\ k\in B}\frac{\mu^{k}}{k!},$$

with the exponential function. The terms are nonnegative, so the sum over the countable index set $\mathbb{N}_0\cap B$ is well-defined independently of ordering as the supremum of its finite partial sums; for the full index set $\mathbb{N}_0$ this unordered sum agrees with the limit of the partial sums of the series $\sum_{k=0}^{\infty}\mu^{k}/k!$, since those partial sums are nondecreasing and every finite subset of $\mathbb{N}_0$ is contained in an initial segment.

$P_\mu$ is a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Countable additivity: if $B_1,B_2,\dots$ are pairwise disjoint Borel sets with union $B$, then every finite subset of $\mathbb{N}_0\cap B$ meets only finitely many of the $B_j$, so the supremum of the finite partial sums over $\mathbb{N}_0\cap B$ equals the sum over $j$ of the suprema over the blocks $\mathbb{N}_0\cap B_j$. Total mass: the sum $\sum_{k\in\mathbb{N}_0}\mu^{k}/k!$ is exactly the defining series $\sum_{k=0}^{\infty}\mu^{k}/k!$ of $\exp(\mu)$ from The Real Exponential Function, so

$$P_\mu(\mathbb{R})=\exp(-\mu)\exp(\mu)=\exp(0)=1$$

by Basic Properties of the Exponential Function. In particular, for $\mu=0$ only the $k=0$ term is nonzero, so $P_0(B)=1$ if $0\in B$ and $P_0(B)=0$ otherwise; we call $P_0$ the unit mass at $0$.

A random variable has the Poisson distribution with parameter $\mu$ if its distribution equals $P_\mu$; such a variable lies in $\mathbb{N}_0$ with probability $1$, since $P_\mu(\mathbb{R}\setminus\mathbb{N}_0)=0$.