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I'm trying to prove that Prokhorov metric satisfies triangle inequality, i.e.,

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

Could you verify if my attempt is fine?

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Analyst
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  • The proof should be part of the question. Then the subsequent answers are "no, here is a subtle flaw..." – Randall Apr 22 '22 at 17:13
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    @Randall I observe that other people tend to comment rather than write an answer. I guess it's partly because their answer is quite short. It makes my question "unanswered" even though I already got one. That's why I post my proof separately. This allows me to subsequently remove this question from unanswered list. If other people post an answer, of course I will accept theirs. – Analyst Apr 22 '22 at 17:17

1 Answers1

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Fix $\mu, \nu, \lambda \in \mathcal P(X)$. Let $$ \mathcal A := \left\{ \alpha + \beta \,\middle\vert\, \alpha,\beta>0 \text{ such that } \begin{align*} \mu(A) \leq \nu\left(A_{\alpha}\right)+\alpha \\ \nu(A) \leq \mu \left(A_{\alpha}\right)+\alpha \\ \nu(A) \leq \lambda\left(A_{\beta}\right)+\beta \\ \lambda(A) \leq \nu \left(A_{\beta}\right)+\beta \end{align*} \quad \forall A \in \mathcal{B}(X) \right\}. $$ and $$ \mathcal A' := \left\{ \alpha>0 \,\middle\vert\, \begin{align*} \mu(A) \leq \lambda\left(A_{\alpha}\right)+\alpha \\ \lambda(A) \leq \mu \left(A_{\alpha}\right)+\alpha \end{align*} \quad \forall A \in \mathcal{B}(X) \right\}. $$

We have $\inf O_1+\inf O_2 \ge \inf (O_1+O_2)$ for all $O_1, O_2 \subseteq \mathbb R$, so $d_{P}(\mu, \nu) + d_{P}(\nu, \lambda) \ge \inf \mathcal A$. It remains to prove that $\inf \mathcal A \ge \inf \mathcal A'$. Again, it suffices to prove that $\mathcal A \subseteq \mathcal A'$.

Let $\alpha,\beta>0$. Fix $x \in (A_\alpha)_{\beta}$. We have $d(x,y') \le d(x,y)+d(y,y')$ for all $y\in A_\alpha$ and $y' \in A$. Then for all $y \in \mathcal A_\alpha$, \begin{align} \inf_{y' \in A}d(x,y') &\le d(x,y) + \inf_{y'\in A} d(y, y') \\ &= d(x,y) +d(y,A) \\ &\le d(x,y)+\alpha \end{align} Then \begin{align} \inf_{y' \in A}d(x,y') &\le\inf_{y \in \mathcal A_\alpha} d(x,y) + \alpha \\ &= d(x, \mathcal A_\alpha)+\alpha \\ &\le \beta+\alpha \end{align} Hence $x \in A_{\alpha+\beta}$ and thus $(A_\alpha)_{\beta} \subseteq A_{\alpha+\beta}$.

Assume $\alpha+\beta \in \mathcal A$. Then $\mu(A) \leq \nu\left(A_{\alpha}\right)+\alpha$ and $\nu(A_\alpha) \leq \lambda\left((A_\alpha)_{\beta}\right)+\beta \le \lambda(A_{\alpha+\beta})+\beta$ for all $A \in \mathcal B(X)$. It follows that $\mu(A) \le \lambda(A_{\alpha+\beta})+\alpha+\beta$ for all $A \in \mathcal B(X)$. Similarly, we have $\lambda(A) \le \mu(A_{\alpha+\beta})+\alpha+\beta$ for all $A \in \mathcal B(X)$. Hence $\alpha+\beta \in \mathcal A'$. This completes the proof.

Analyst
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