2
$\begingroup$

I'm reading about Prokhorov metric, i.e.,

Let $(X, d)$ be a metric space and $\mathcal{P} :=\mathcal{P}(X)$ the set all Borel probability measures on $X$. Let $$ d_{P}(\mu, \nu) := \inf \left\{ \alpha>0 \,\middle\vert\, \begin{align*} \mu(A) \leq \nu\left(A_{\alpha}\right)+\alpha \\ \nu(A) \leq \mu \left(A_{\alpha}\right)+\alpha \end{align*} \quad \forall A \in \mathcal{B}(X) \right\} \quad \forall \mu, \nu \in \mathcal{P}, $$ where $A_{\alpha} := \{x \mid d(x, A)<\alpha\}$ and $d(x, A)=\inf \{d(x, a) \mid a \in A\}$.

Below is the proof that $d_P$ is indeed a metric.

  • Any $\alpha \geq 1$ is in the set of the defining formula of $d_{P}$, so the infimum is well defined.

  • Clearly, $d_{P}(\mu, \nu) \geq 0$ and $d_{P}(\mu, \nu)=d_{P}(\nu, \mu)$ for all $\mu, \nu \in \mathcal{P}$.

  • Let $\mu \in \mathcal{P}$. For every $A \in \mathcal{B}(X)$ and $\alpha>0$, $A \subseteq A_{\alpha}$, so $\mu(A) \leq \mu\left(A_{\alpha}\right)+\alpha$. Hence $d_{P}(\mu, \mu) \leq \alpha$ and thus $d_{P}(\mu, \mu)=0$.

  • If $d_{P}(\mu, \nu)=0$, then there is a sequence $\alpha_{n} \downarrow 0$ such that $\mu(A) \leq \nu\left(A_{\alpha_{n}}\right)+\alpha_{n}$ and $\nu(A) \leq \mu\left(A_{\alpha_{n}}\right)+\alpha_{n}$ for all $n$. As $\bar{A}=\bigcap_{n} A_{\alpha_{n}}$, it follows that $\mu(A) \leq \nu(\bar{A})$ and $\nu(A) \leq \mu(\bar{A})$. In particular, $\mu(A)=\nu(A)$ for all closed sets $A$ and therefore $\mu=\nu$ by inner regularity.

  • The triangle inequality is proved here.


It seems to me this metric can be extended to all finite non-negative Borel measures. For the defining formula of $d_P$ to be well-defined, we just replace $\alpha \ge 1$ by $\alpha = \max \{\mu(X), \nu(X)\}$.

Could you confirm if my observation is correct?

$\endgroup$

1 Answer 1

1
+100
$\begingroup$

I just check all the proofs, which only use finiteness for finite many measures.

So the answer is yes, you're right.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.