# Variational distance of product of distributions

Let $F(\bar{x})=\prod_{i=1}^{n}f(x_i)$ and $G(\bar{x})=\prod_{i=1}^{n}g(x_i)$, where $f(x)$ and $g(x)$ are probability density functions, and $\bar{x}=(x_1,\ldots,x_n)$. The variational distance between $F$ and $G$ is: $$V(F,G)=\int |F(\bar{x})-G(\bar{x})|d\bar{x}$$ Can we write it in terms of the variational distance between $f$ and $g$?

I know we could do such if it was KL divergence: $$D_{KL}(F,G)=n D_{KL}(f,g).$$ Do we have such a simplification for variational distance as well?

• No, we don't. The TV distance simply doesn't tensorize well. A saving grace, however, is the fact that TV is easily bounded by many $f$-divergences (e.g., Pinsker's inequality says $V \le \sqrt{2 D_{KL}}.$), which is many times sufficient to solve the problem at hand. Alternately one has the choice to 'metrise' the problems using some other divergence. KL, $\chi^2,$ Jensen-Shannon and Hellinger are popular choices, with the last two both being symmetric in their arguments, and being 'close to' actual norms, if that is important. – stochasticboy321 Sep 14 '18 at 2:49