What puzzled me in reading the abstract is the question of how are these distributions given to us? I somehow missed the natural answer: These are product distributions over 0-1 variables, and so they are described by their margianls. So the question studied here is, given $(p_1,...,p_n)$ and $(q_1,...,q_n)$, compute (or approximate) the difference
$$\sum_{I\subseteq[n]} |(\prod_{i\in I}p_i)\cdot(\prod_{i\in[n]-I}(1-p_i)) -(\prod_{i\in I}q_i)\cdot(\prod_{i\in[n]-I}(1-q_i))|$$Interestingly, this probalem is $\#P$-complete, whereas it has a FPTAS.
Both claims rely on the fact that if the expression is positive then it can be lower-bounded in terms of the input length. Since the hardness claim is more suprrising, it is adequate to hint that it is proved by reduction to the problem of determining, when given $u_1,...,u_n$ and $v$, the number of $I$'s that satisfy
$$(\prod_{i\in I}u_i)\cdot(\prod_{i\in[n]-I}(1-u_i)) = v,$$which can be realted to a subset sum problem (by devising by $\prod_{i\in[n]} (1-u_i)$ and taking logs).
The approximation algorithm works by using (multiplicative) bucketing (of contributions to an expression corresponding to the TVD), and approximating the number of elements in each bucket.
Total variation distance (TV distance) is a fundamental notion of distance between probability distributions. In this work, we introduce and study the computational problem of determining the TV distance between two product distributions over the domain {0,1}n. We establish the following results.
See arxiv 2206.07209.