Let $\mathbb{N}$ be the set of \reftext{def:natural-numbers-2026a}{natural numbers} and write $\mathbb{N}_0=\mathbb{N}\cup\{0\}$ for the set of \textbf{nonnegative integers}; let $\mathbb{R}$ be the set of \reftext{def:real-numbers-c54-2026c}{real numbers} and $\mathcal{B}(\mathbb{R})$ the \reftext{def:borel-sigma-algebra-real-line-2026a}{Borel $\sigma$-algebra}. We use the \reftext{def:factorial-natural-number-2026a}{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 \textbf{Poisson distribution} with parameter $\mu$ is the function

with the \reftext{def:exponential-function-real-2026a}{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 \reftext{def:measure-measure-space-2026a}{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 \ref{def:exponential-function-real-2026a}, so

by \ref{thm:exponential-properties-2026a}. 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 \textbf{unit mass at $0$}.

A random variable has the \textbf{Poisson distribution with parameter $\mu$} if its \reftext{def:distribution-cdf-random-variable-2026a}{distribution} equals $P_\mu$; such a variable lies in $\mathbb{N}_0$ with probability $1$, since $P_\mu(\mathbb{R}\setminus\mathbb{N}_0)=0$.