# 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 $\pi (\mu, \nu) \le 1$.

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.