Lévy–Prokhorov metric

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

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

${\displaystyle A^{\varepsilon }:=\{p\in M~|~\exists q\in A,\ d(p,q)<\varepsilon \}=\bigcup _{p\in A}B_{\varepsilon }(p).}$

where ${\displaystyle B_{\varepsilon }(p)}$ is the open ball of radius ${\displaystyle \varepsilon }$ centered at ${\displaystyle p}$.

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

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

For probability measures clearly ${\displaystyle \pi (\mu ,\nu )\leq 1}$.

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

• If ${\displaystyle (M,d)}$ is separable, convergence of measures in the Lévy–Prokhorov metric is equivalent to weak convergence of measures. Thus, ${\displaystyle \pi }$ is a metrization of the topology of weak convergence on ${\displaystyle {\mathcal {P}}(M)}$.
• The metric space ${\displaystyle \left({\mathcal {P}}(M),\pi \right)}$ is separable if and only if ${\displaystyle (M,d)}$ is separable.
• If ${\displaystyle \left({\mathcal {P}}(M),\pi \right)}$ is complete then ${\displaystyle (M,d)}$ is complete. If all the measures in ${\displaystyle {\mathcal {P}}(M)}$ have separable support, then the converse implication also holds: if ${\displaystyle (M,d)}$ is complete then ${\displaystyle \left({\mathcal {P}}(M),\pi \right)}$ is complete. In particular, this is the case if ${\displaystyle (M,d)}$ is separable.
• If ${\displaystyle (M,d)}$ is separable and complete, a subset ${\displaystyle {\mathcal {K}}\subseteq {\mathcal {P}}(M)}$ is relatively compact if and only if its ${\displaystyle \pi }$-closure is ${\displaystyle \pi }$-compact.
• If ${\displaystyle (M,d)}$ is separable, then ${\displaystyle \pi (\mu ,\nu )=\inf\{\alpha (X,Y):{\text{Law}}(X)=\mu ,{\text{Law}}(Y)=\nu \}}$, where ${\displaystyle \alpha (X,Y)=\inf\{\varepsilon >0:\mathbb {P} (d(X,Y)>\varepsilon )\leq \varepsilon \}}$ is the Ky Fan metric.[1][2]

Let ${\displaystyle (M,d)}$ be separable. Then

• ${\displaystyle \pi (\mu ,\nu )\leq \delta (\mu ,\nu )}$, where ${\displaystyle \delta (\mu ,\nu )}$ is the total variation distance of probability measures[3]
• ${\displaystyle \pi (\mu ,\nu )^{2}\leq W_{p}(\mu ,\nu )^{p}}$, where ${\displaystyle W_{p}}$ is the Wasserstein metric with ${\displaystyle p\geq 1}$ and ${\displaystyle \mu ,\nu }$ have finite ${\displaystyle p}$th moment.[4]

• Lévy metric
• Prokhorov's theorem
• Tightness of measures
• Weak convergence of measures
• Wasserstein metric
• Radon distance
• Total variation distance of probability measures

• Billingsley, Patrick (1999). Convergence of Probability Measures. John Wiley & Sons, Inc., New York. ISBN 0-471-19745-9. OCLC 41238534.
• Zolotarev, V.M. (2001) [1994], "Lévy–Prokhorov metric", Encyclopedia of Mathematics, EMS Press
• Dudley, R.M. (1989). Real analysis and probability. Pacific Grove, Calif. : Wadsworth & Brooks/Cole. ISBN 0-534-10050-3.
• Račev, Svetlozar T. (1991). Probability metrics and the stability of stochastic models. Chichester [u.a.] : Wiley. ISBN 0-471-92877-1.