3
$\begingroup$

Let $(\mu_0,\mu_1)$ be a vector of probability measures on $\mathbb R$ that are of finite first moment, i.e.

$$\int_{\mathbb{R}}|x|\mu_i(dx)~<~+\infty \mbox{ for } i=0,1$$

and increasing in convex order, i.e.

$$\int_{\mathbb{R}}\lambda(x)\mu_0(dx)~\le~\int_{\mathbb{R}}\lambda(x)\mu_1(dx) \mbox{ for all convex function } \lambda \mbox{ with linear growth}.$$

It is clear that (by either Strassen's theorem or Kellerer's theoerm), there exists a martingale measure $\pi$ on $\mathbb R^2$ with the given marginal distributions $\mu_0$ and $\mu_1$, i.e.

$$\pi^i=\mu_i \mbox{ for } i=0,1~~~ \mbox{ and } ~~~\int_{\mathbb R}y\pi_x(dy)=x \mbox{ for } \mu_0\mbox{ - a.e. } x\in\mathbb R,$$

where $\pi^i$ denotes the marginal distribtution of the $i$th component and $\pi_x$ denotes the disintegration of $\pi$ w.r.t. $\pi^0=\mu_0$.

Fix the martingale measure $\pi$. We consider another vector of probability measures $(\nu_0,\nu_1)$ which are of finite first moment and increasing in convex order. My question is the following:

Assume that $\mathcal W_1(\mu_i,\nu_i)<\varepsilon$ for $i=0,1$, where $\mathcal W_1$ is the Wasserstein distance. Could we find a martingale measure $\gamma$ with marginal distributions $\nu_0$ and $\nu_1$ s.t. $\mathcal W_1(\pi,\gamma)<\alpha(\varepsilon)$ for some function $\alpha$ satisfying $\lim_{x\to 0}\alpha(x)=0$.

See e.g. for the definition of Wasserstein distance

https://en.wikipedia.org/wiki/Wasserstein_metric

Thanks a lot!

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.