Wasserstein distance and put function

Question

Let $\mathcal P$ be the set of probability distributions on $\mathbb R$ of finite first order, i.e. $\mu\in\mathcal P$ if

$$\int_{\mathbb R} |t|\mu(dt)<\infty.$$

For $\mu\in\mathcal P$, define its put function $P_\mu:\mathbb R\to\mathbb R_+$ by

$$P_\mu(x):=\int_{\mathbb R} (x-t)^+\mu(dt).$$

Can we control the Wasserstein distance $W$ (of order $1$) in terms of the put function? Namely, does there exist some function $G$ (as explicit as possible) such that for all $\mu,\nu\in\mathcal P$

$$W(\mu,\nu)\le G\big(P_\mu,P_\nu\big)?$$

Of course $G\big(P_\mu,P_\mu\big)=0$ should be expected. I tried with the dual formulation of $W$, i.e.

$$W(\mu,\nu)=\int_{\mathbb R} |F_\mu(x)-F_\nu(x)|dx,$$

where $F_\mu$ denotes the cumulative function of $\mu$. Using further

$$xF_\mu(x) = P_\mu(x) +\int_{(-\infty,x]}t\mu(dt),$$

one has

$$W(\mu,\nu)=\int_{\mathbb R} \left|\frac{\big(P_\mu(x)-P_\nu(x)\big) +\left(\int_{(-\infty,x]}t\mu(dt)-\int_{(-\infty,x]}t\nu(dt)\right)}{x}\right|dx.$$

Is there any idea to deal with the above expression?

Answer 1

$\newcommand\R{\Bbb R}\newcommand\LS{\mathsf{LS}}$For real $x$ and $y$ such that $x<y$, $$\begin{aligned} P_\mu(y)-P_\mu(x)&=\int_\R[(y-t)_+-(x-t)_+]\mu(dt) \\ &=\int_\R[(y-x)\,1(t\le x)+(y-t)\,1(x<t<y)]\mu(dt) \\ &=(y-x)\mu((-\infty,x])+O((y-x)\mu((x,y))). \end{aligned}$$ So, the right derivative of the continuous function $P_\mu$ at $x$ is $$(P_\mu)'_+(x)=\mu((-\infty,x]).$$ Similarly, the left derivative of $P_\mu$ at $x$ is $$(P_\mu)'_-(x)=\mu((-\infty,x)).$$ Let now $$G(P_\mu,P_\nu):=W(\LS((P_\mu)'_+),\LS((P_\nu)'_+)),$$ where $\LS((P_\mu)'_+)$ is the Lebesgue--Stiltjes measure corresponding to the right-continuous nondecreasing function $(P_\mu)'_+$.

Then $$W(\mu,\nu)=G\big(P_\mu,P_\nu\big),$$ so that $G\big(P_\mu,P_\nu\big)$ is the best upper bound on $W(\mu,\nu)$ in terms of $P_\mu$ and $P_\nu.\quad$ :-)