Find $\inf_{P_{X_1,X_2}}P_{X_1,X_2}(\|X_1-X_2\| > 2\alpha)$ , where $\alpha > 0$ and inf is over couplings

Question

Let $\mathcal X$ be a seperable Banach space with norm $\|\cdot\|$, and let $X_1$ and $X_2$ be random vectors on $\mathcal X$ with finite means.

Question. Given $\alpha > 0$, what is value of, or an alternative expression for $$ h(\alpha):=\inf_{P_{X_1,X_2}}P_{X_1,X_2}(\|X_1-X_2\| > 2\alpha), $$ where the infimum is taken over all joint distributions $P_{X_1,X_2}$ of $X_1$ and $X_2$ ?

Motivation. It's well-known that $\text{TV}(X_1,X_2) = \inf_{P_{X_1,X_2}}P_{X_1,X_2}(X_1 \ne X_2)$. Thus, $h(0)=\text{TV}(X_1,X_2)$ in particular.

Conjecture. $h(\alpha) =\inf_{\|z\| \le \alpha}\text{TV}(X_1-z,X_2+z)$.

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Edit: Proof of weaker form of conjecture: an upper bound

In the comments to this question and answers, many users have pointed out counter-examples to my Conjecture. Here, I'll settle to proof a weaker version: an inequality. For simplicity of notation, let $P_k$ be the distribution of $X_k$. Viz,

Theorem. $h(\alpha) \le \inf_{\|z\| \le \alpha}TV(P_1-z,P_2+z)$.

User Iosif has established that

$h(\alpha) = \sup\{P_1(U)-P_2(U^{2\alpha})\mid U \subseteq \mathcal X\text{ open}\},$

where $U^{\delta} = \{x_1 \in \mathcal X \mid d(x_1,U) \le \delta\}$ is the $\delta$-neighborhood of $U$, and $d(x_1,U) := \inf_{x_2 \in U}\|x_1-x_2\|$ is the distance of $x_1$ from the set $U$. I'll use this to proof the above theorem (the TV upper bound).

Now, for every $z \in \mathcal X$, we may translate all the open sets $U$ in the above formula without changing it. Indeed the invariance $\{U-z \mid U \subseteq \mathcal X\text{ open}\} = \{U \mid U \subseteq X\text{ open}\}$ is trivial to show. Thus, $$ h(\alpha) = \sup\{P_1(U-z)-P_2((U-z)^{2\alpha})\mid U \subseteq \mathcal X\text{ open}\}. $$ On the other hand, it is clear that $U+z \subseteq (U-z)^{2\alpha}$ whenever $\|z\| \le \alpha$. Indeed, $$ x=u + z \in U + z \implies d(x,U-z) = d(u, U-2z) \le d(u,u-2z) = 2\|z\| \le 2\alpha. $$ Thus $P_1(U-z)-P_2((U-z)^{2\alpha}) \le P_1(U-z)-P_2(U+z)$ $\forall$ open $U \subseteq \mathcal X$ and $\|z\| \le \alpha$.

$$ \therefore h(\alpha) \le \inf_{\|z\| \le \alpha}\sup\{P_1(U-z)-P_2(U+z)\mid U \subseteq \mathcal X\text{ open}\} = \inf_{\|z\| \le \alpha}TV(P_1-z,P_2+z). $$ This completes the proof of the theorem.

Answer 1

Let $\mathcal X$ be a separable Banach space. Let $\mu$ and $\nu$ be the distributions of $X_1$ and $X_2$, respectively. Then, by Theorem 11 (pp. 436-437) of Strassen (used with $\omega=\{(x_1,x_2)\in\mathcal X\times\mathcal X\colon\|x_1-x_2\|\le2\alpha\}$), $$h(\alpha)=\sup\{\nu(U)-\mu(U^{2\alpha})\colon U\subseteq\mathcal X, U\text{ is open}\}, $$ where $U^\delta$ is the $\delta$-neighborhood of $U$.

In particular, it follows that indeed $h(0)=TV(\mu,\nu)$.

Answer 2

As Iosif wrote, the conjecture does not hold. Suppose $\alpha=1$ and $X_1$ takes the values 0,1,2 with equal probability and $X_2$ takes the values 0,2 with equal probability. Then $h(1)=0$ but translates of these variables have TV distance at least $ 1/3$.