Total variation distance between pairs when one variable is independent of the rest Let $\Delta(\cdot, \cdot)$ denote total variation distance. Let $X, Y, U, V$ be random variables, where $U$ is independent of $(X, Y, V)$. Does this imply $\Delta((X, Y), (U, Y)) \leq \Delta((X, Y), (U, V))$? For simplicity, assume all the random variables are discrete.
Surely there is a simple proof or a simple counterexample, but so far I can't find either. If we drop the independence assumption, it's easy to find counterexamples, e.g., sample $A \in \{0, 1\}$ uniformly at random and let $X = Y = A$ and $U = V = 1 - A$.
 A: As mentioned in the comments, to remove some notational ambiguity I am rephrasing your question as

Let $p$ be a probability distribution on some discrete space $\mathcal{X}\times\mathcal{Y}$, with marginals $p_X$ and $p_Y$; and let $u,v$ be two probability distributions on $\mathcal{X}$ and $\mathcal{Y}$ respectively. Is it always the case that $\Delta(p, u\otimes p_Y) \leq\Delta(p, u\otimes v)$
?

The answer is no. Take for instance $\mathcal{X}=\mathcal{Y}=\{0,1\}$, and $p,u,v$ defined as follows:
$$
p(0,0) = p(1,1)=0, \quad p(0,1)=p(1,0)=\frac{1}{2}
$$
and
$$
u(0)=\frac{1}{10}, \;u(1)=\frac{9}{10} \qquad\text{and}\qquad v(0)=\frac{9}{10}, \;v(1)=\frac{1}{10}
$$
Note that the marginal $p_Y$ of $p$ is uniform on $\{0,1\}$.
Then, we can compute
$$
2\Delta(p, u\otimes p_Y) = |\frac{1}{10}\cdot\frac{1}{2}-0|+|\frac{1}{10}\cdot\frac{1}{2}-\frac{1}{2}|+|\frac{9}{10}\cdot\frac{1}{2}-\frac{1}{2}|+|\frac{9}{10}\cdot\frac{1}{2}-0|=1
$$
while
$$
2\Delta(p, u\otimes v) = |\frac{1}{10}\cdot\frac{9}{10}-0|+|\frac{1}{10}\cdot\frac{1}{10}-\frac{1}{2}|+|\frac{9}{10}\cdot\frac{9}{10}-\frac{1}{2}|+|\frac{9}{10}\cdot\frac{1}{10}-0|=\frac{98}{100}
$$
so $\boxed{\Delta(p, u\otimes p_Y) > \Delta(p, u\otimes v)}$.
