# Lévy–Prokhorov metric

In mathematics, the Lévy–Prokhorov metric (sometimes known just as the Prokhorov metric) is a metric (i.e., a definition of distance) on the collection of probability measures on a given metric space. It is named after the French mathematician Paul Lévy and the Soviet mathematician Yuri Vasilyevich Prokhorov; Prokhorov introduced it in 1956 as a generalization of the earlier Lévy metric.

## Definition

Let $(M,d)$ be a metric space with its Borel sigma algebra ${\mathcal {B}}(M)$ . Let ${\mathcal {P}}(M)$ denote the collection of all probability measures on the measurable space $(M,{\mathcal {B}}(M))$ .

For a subset $A\subseteq M$ , define the ε-neighborhood of $A$ by

$A^{\varepsilon }:=\{p\in M~|~\exists q\in A,\ d(p,q)<\varepsilon \}=\bigcup _{p\in A}B_{\varepsilon }(p).$ where $B_{\varepsilon }(p)$ is the open ball of radius $\varepsilon$ centered at $p$ .

The Lévy–Prokhorov metric $\pi :{\mathcal {P}}(M)^{2}\to [0,+\infty )$ is defined by setting the distance between two probability measures $\mu$ and $\nu$ to be

$\pi (\mu ,\nu ):=\inf \left\{\varepsilon >0~|~\mu (A)\leq \nu (A^{\varepsilon })+\varepsilon \ {\text{and}}\ \nu (A)\leq \mu (A^{\varepsilon })+\varepsilon \ {\text{for all}}\ A\in {\mathcal {B}}(M)\right\}.$ For probability measures clearly $\pi (\mu ,\nu )\leq 1$ .

Some authors omit one of the two inequalities or choose only open or closed $A$ ; either inequality implies the other, and $({\bar {A}})^{\varepsilon }=A^{\varepsilon }$ , but restricting to open sets may change the metric so defined (if $M$ is not Polish).

## Properties

• If $(M,d)$ is separable, convergence of measures in the Lévy–Prokhorov metric is equivalent to weak convergence of measures. Thus, $\pi$ is a metrization of the topology of weak convergence on ${\mathcal {P}}(M)$ .
• The metric space $\left({\mathcal {P}}(M),\pi \right)$ is separable if and only if $(M,d)$ is separable.
• If $\left({\mathcal {P}}(M),\pi \right)$ is complete then $(M,d)$ is complete. If all the measures in ${\mathcal {P}}(M)$ have separable support, then the converse implication also holds: if $(M,d)$ is complete then $\left({\mathcal {P}}(M),\pi \right)$ is complete.
• If $(M,d)$ is separable and complete, a subset ${\mathcal {K}}\subseteq {\mathcal {P}}(M)$ is relatively compact if and only if its $\pi$ -closure is $\pi$ -compact.