Compute explicitly Lévy–Prokhorov metric for $2$ finite Dirac measures

Question

Let $(X, d)$ be a metric space and $\mathcal{M} :=\mathcal{M}(X)$ the space all non-negative finite Borel measures on $X$. The Prokhorov metric $d_P$ on $\mathcal{M}$ is defined by $$ d_{P}(\mu, \nu) := \inf \left\{ \alpha>0 \,\middle\vert\, \begin{align*} \mu(A) \leq \nu\left(A_{\alpha}\right)+\alpha \\ \nu(A) \leq \mu \left(A_{\alpha}\right)+\alpha \end{align*} \quad \forall A \in \mathcal{B}(X) \right\}, $$ with $A_{\alpha} := \{x \mid d(x, A)<\alpha\}$ and $d(x, A) := \inf \{d(x, a) \mid a \in A\}$.

Previously, I proved that $d_P$ is indeed a metric. In my lecture note, $d_P$ has been computed for Dirac probability measures, i.e., $$ d_P (\delta_x, \delta_y) = \min\{d(x, y), 1\} \quad \forall x,y \in X. $$

Now I try to compute it for finite Dirac measures, i.e.,

Let $\mu := a \delta_x$ and $\nu := b \delta_y$ with $a,b \ge 0$ and $x,y \in X$. Compute $d_P(\mu, \nu)$.

Could you have a check on my attempt?

I post my proof separately as below answer. If other people post an answer, of course I will happily accept theirs. Otherwise, this allows me to subsequently remove this question from unanswered list.

Answer 1

1. Let $A = \{x\}$. Then $\mu(A)=\mu(A_\alpha) = a$. Also, $\nu(A) = b1_{x=y}$ and $\nu(A_\alpha) = b1_{\{\alpha>d(x,y)\}}$. If $\alpha > d(x, y)$, the conditions become $\alpha \ge \max \{a-b, b1_{x=y}-a\}$. If $\alpha \le d(x, y)$, the conditions become $\alpha \ge \max \{a, b1_{x=y}-a\}$.

2. Let $A = \{y\}$. If $\alpha > d(x, y)$, the conditions become $\alpha \ge \max \{b-a, a1_{x=y}-b\}$. If $\alpha \le d(x, y)$, the conditions become $\alpha \ge \max \{b, a1_{x=y}-b\}$.

3. Let $A = \{x, y\}$. Then $\alpha \ge \max \{ a-b, b-a \} = |a-b|$.

Clearly, $|a-b| \ge \max \{b1_{x=y}-a, a1_{x=y}-b\}$. So the combination of above cases reduces to

1. Let $A = \{x\}$. If $\alpha > d(x, y)$, the conditions become $\alpha \ge |a-b|$. If $\alpha \le d(x, y)$, the conditions become $\alpha \ge \max \{a, |a-b|\}$.

2. Let $A = \{y\}$. If $\alpha > d(x, y)$, the conditions become $\alpha \ge |a-b|$. If $\alpha \le d(x, y)$, the conditions become $\alpha \ge \max \{b, |a-b|\}$.

3. Let $A = \{x, y\}$. Then $\alpha \ge |a-b|$.

So the set of eligible $\alpha$ on which we take the infimum is $$ \mathcal A := \{ \alpha > d(x,y), \alpha \ge |a-b|\} \bigcup \{ \alpha \le d(x,y), \alpha \ge \max \{a,b,|a-b|\}\}. $$

Notice that $a,b \ge0$, so $\alpha \ge \max \{a,b,|a-b|\} \iff \alpha \ge \{a,b\}$. Then $$ \mathcal A = \{ \alpha > d(x,y), \alpha \ge |a-b|\} \bigcup \{ \alpha \le d(x,y), \alpha \ge \max \{a,b\} \}. $$

• If $\max \{a,b\} \le d(x,y)$, then $|a-b| \le d(x,y)$. So $$ \begin{align} \mathcal A &= \{ \alpha > d(x,y)\} \bigcup \{ \max \{a,b\} \le \alpha \le d(x,y)\} \\ &= \{ \alpha > d(x,y)\}. \end{align} $$ Then $d_P(\mu, \nu) = \inf \mathcal A = \max\{a,b\}$.

• If $\max \{a,b\} > d(x,y)$, then $$ \begin{align} \mathcal A &= \{ \alpha > d(x,y), \alpha \ge |a-b| \}. \end{align} $$ Then $d_P(\mu, \nu) = \inf \mathcal A = \max \{d(x,y), |a-b|\}$.