Total variation distance equality

Question

I'm trying to show that the equality is possible in: $P(X = Y) \le 1 -1/2 \operatorname{DTV} (X,Y)$

where DTV = total variation distance.

For any pair $X,Y$ there exists a pair $X', Y'$ having the same marginal distributions as $X,Y$ s.t. $P(X'=Y') = 1 -1/2 \operatorname{DTV}(X,Y)$

I thought of using the fact that If f is probability density function and $(U, V )$ is a point chosen uniformly at random from the region $\{(u, v) : -\infty < u < \infty,\ 0 \le v \le f(u)\}$ then rndm variable $U$ has probability density function $f$.

Answer 1

Let $\mu$ denote the distribution of $X$ and $\nu$ the distribution of $Y$. Consider the total variation distance $d(\mu,\nu)=\max_B\mu(B)-\nu(B)$, hence $0\leqslant d(\mu,\nu)\leqslant1$ (and $d$ is probably one half of your DTV). There exists a unique maximal positive measure, usually denoted $\mu\wedge\nu$, such that $\mu-(\mu\wedge\nu)$ and $\nu-(\mu\wedge\nu)$ are mutually singular nonnegative measures. Call $s$ the total mass of $\mu\wedge\nu$, then $d(\mu,\nu)=1-s$ is the common total mass of both measures $\mu-(\mu\wedge\nu)$ and $\nu-(\mu\wedge\nu)$. Define the pair $(X',Y')$ as follows.

Let $S$ denote a Bernoulli random variable with parameter $s=\mathrm P(S=1)=1-\mathrm P(S=0)$. Let $U$, $V$ and $W$ denote independent random variables, independent on $S$, with distributions $$ \mathrm P_U=(1-s)^{-1}(\mu-(\mu\wedge\nu)),\quad \mathrm P_V=(1-s)^{-1}(\nu-(\mu\wedge\nu)),\quad \mathrm P_W=s^{-1}(\mu\wedge\nu). $$ Define $$ X'=SW+(1-S)U,\qquad Y'=SW+(1-S)V. $$ In words, if $S=1$, $X'=Y'=W$ while, if $S=0$, $X'=U$ and $Y'=V$.

Then $\mathrm P_U$ and $\mathrm P_V$ are mutually singular hence $\mathrm P(U=V)=0$. Thus, $[X'=Y']=[S=1]$, which implies $\mathrm P(X'=Y')=s=1-d(\mu,\nu)$.

This construction is a classical application of the coupling method and is explained, if I remember correctly, in the first chapter of Lindvall's book.