Let $X,Y$ and $Z$ be random vectors in $\mathbb{R}^d$ defined on $(\Omega,\mathcal{H},\mathsf{P})$ s.t. $Z$ is $\mathcal{F}$-measurable for some $\mathcal{F}\subset\mathcal{H}$. Define a family of random, convex, closed sets: $$ A_t(\omega):=\{x\in \mathbb{R}^d:x^{\top}Z(\omega)\le t\} $$ indexed by $t\in\mathbb{R}$. Is there a family of sets $\mathcal{C}\subset\mathcal{B}(\mathbb{R}^d)$ (independent of $Z$) s.t. for each $t\in \mathbb{R}$, \begin{align} &|\mathsf{P}(X\in A_t\mid\mathcal{F})-\mathsf{P}(Y\in A_t\mid \mathcal{F})| \\ &\qquad\le \operatorname{esssup_{C\in\mathcal{C}}}|\mathsf{P}(X\in C\mid\mathcal{F})-\mathsf{P}(Y\in C\mid \mathcal{F})| \quad\text{a.s.}? \end{align}

In particular, does the inequality hold for $\mathcal{C}=\{x,z\in\mathbb{R}^d,t\in\mathbb{R}:x^{\top}z\le t\}$?

The case $d=1$ is relatively simple. Assume, for simplicity, that $Z>0$ a.s. and let $P_X(A,\cdot)$ be the Markov kernel corresponding to $\mathsf{P}(X\in A\mid \mathcal{F})$ ($P_Y(A,\cdot)$ is defined similarly). Then \begin{align} &|\mathsf{P}(X\in A_t\mid\mathcal{F})-\mathsf{P}(Y\in A_t\mid \mathcal{F})| \\ &\qquad =|P_X((-\infty,w])-P_Y((-\infty,w])| \quad\text{a.s.}, \end{align} where $w=t/Z$ and $w$ is $\mathcal{F}$-measurable. Approximating $w$ from above by a sequence of step functions and taking the limit ($\because$ the map $t\mapsto |P_X((-\infty,t],\omega)-P_Y((-\infty,t],\omega)|$ is right-continuous) we get \begin{align} &|\mathsf{P}(X\in A_t\mid\mathcal{F})-\mathsf{P}(Y\in A_t\mid \mathcal{F})| \\ &\qquad\le \operatorname{esssup_{C\in\mathcal{C}}}|\mathsf{P}(X\in C\mid\mathcal{F})-\mathsf{P}(Y\in C\mid \mathcal{F})| \quad\text{a.s.}, \end{align} where $\mathcal{C}=\{(-\infty,t]:t\in\mathbb{R}\}$. However, I don't see a straightforward way to generalize this for $d>1$.

  • $\begingroup$ I guess I'm missing something, but why doesn't $\mathcal C = \{ A_t \}$ get the job done? $\endgroup$ Jul 9, 2018 at 18:53
  • $\begingroup$ @ChristianRemling $A_t$ is random and I'm looking for nonrandom sets that are independent of $Z$. $\endgroup$
    – d.k.o.
    Jul 9, 2018 at 19:06


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