Does strong stochastic ordering exist?

Question

For two probability measure $\mu$ and $\nu$ on $\mathbb{R}$, we call $\mu$ is stochastically smaller than $\nu$ (i.e., $\mu\leq\nu$) , if $\int f \, d\mu\leq\int f \, d\nu$ for any nonnegative bounded increasing measurable function $f:\mathbb{R}\to\mathbb{R}$.

There is a fact that $\mu\leq\nu$ is equivalent to $F_\mu(x)\geq F_\nu(x)$ for all $x\in\mathbb{R}$, where $F_\mu(x)$ and $F_\nu(x)$ are the cumulative distribution functions of $\mu$ and $\nu$ respectively.

Another fact is that, $\mu\leq\nu$ is equivalent to $\int f \, d\mu\leq\int f \, d\nu$ for any nonnegative bounded increasing measurable function $f:\mathbb{R}\to\mathbb{R}$ with Lipschitz constant less or equal to $1.$

I'm curious about whether there is a strong order in this ordered space. That is to say, can we find some topology in probability space such that, there exists two probability measure $\mu$ and $\nu$, and their open neighborhood $\mu\in U$, $\nu\in V$ in this topology such that, $\tilde{\mu}\leq \tilde{\nu}$, for any $\tilde{\mu}\in U$ and $\tilde{\nu}\in V$.

I guess we cannot find such open sets in weak convergence topology, or Wasserstein metric $W_p$, or total variation distance topology. I cannot strictly prove it. The intuition tells me that for any $\mu\leq\nu$, perturbate the distribution function of $\mu$ and $\nu$ such that it changes little near $-\infty$ will cause contradiction.

The same question also exists for other partial order in the probability space.

Any advices will be greatly appreciated.

Answer 1

You could define a distance on probability measures by the smallest $c$ such that there exists a coupling giving mass $1$ to a $c$-neighbourhood of the diagonal. Many pairs would be infinite distance apart (you can turn it into a finite distance by taking the inf with $1$ if you want) but it would have the property you're looking for if I understood the question correctly.

Answer 2

First, for the discrete topology, $U := \{\mu\}$ and $V := \{\nu\}$ are open and have the property wanted. Surely this is excluded. For the weak convergence topology, the Wasserstein metric and the stronger total variation topology its always impossible to find open neighborhoods $U$ and $V$ with the desired property.

It suffices to consider the total variation topology. For any neighborhood $U$ of $\mu$ there is $\tilde \mu \in U$ with $\tilde \mu \not\leq \nu$. Of course we assume that $\mu \not= \nu$.

For let $\mu = \mu_1 + \mu_2$ with $\mu_1$ discrete and $\mu_2$ continuous (i.e. with continuous distribution function). Let $\epsilon$ be the distance of $\mu$ and the complement $U^c$ of $U$. Of course $\epsilon > 0$. Now if $\mu_1 > 0$ there is $x \in \mathbb{R}$ with $0 < \mu_1(\{x\})$. Let $\tilde \epsilon := \min\{\epsilon/3,\mu_1(\{x\}\}$. Assume that $\nu((-\infty,y]) > 1-\epsilon$. Then $\tilde \mu := \mu - \tilde \epsilon \delta_x + \tilde \epsilon \delta_y \in U$, but $\tilde \mu \not\leq \nu$. If $\mu_1 = 0$ we proceed similarly. We cut some part of $\mu$ with mass $\epsilon/3$ and add some $\epsilon/3 \cdot \delta_y$ with large $y$ to $\mu$ as in the former case.