TITLE: How does a statistical divergence change under a Lipschitz push-forward map?
QUESTION [5 upvotes]: Let $\mu, \nu$ be two probability measures on a space $X$ (assume Polish space).
$T: X \rightarrow Y$ is a Lipschitz-map that acts as a push-forward on these measures; let $\mu^\prime = T_{\#\mu}$ and $\nu^\prime = T_{\#\nu}$ be the resulting push-forward measures on the space $Y$. 
Let $d(\mu, \nu)$ denote some distance function between measures. 
What can be said about how $d(\mu, \nu)$ and $d(\mu^\prime, \nu^\prime)$ relate to each other? Especially, how is the relationship a function of the Lipschitz constant of $T$? 
Of course the relationship is likely to be a function of the exact nature of $d()$ (e.g., KL-divergence, or Total-variation or Wassterstein .. ).
I suspect a lot must be known about this type of questions; what would be the right place to look at?

REPLY [3 votes]: If $d(\mu,\nu)$ is taken to be the total variation metric, then Lipschitz and metric properties don't matter. This is due to a "data processing inequality" of sorts: applying a transformation can only make two distributions closer in TV. I'll illustrate this for discrete sets $X,Y$:
$$ d(\mu',\nu') =\sum_{y\in Y}|\mu'(y)-\nu'(y)|
=\sum_{y\in Y}|\sum_{x\in T^{-1}(y)}\mu(x)-\nu(x)|
\le\sum_{x\in X}|\mu(x)-\nu(x)|=d(\mu,\nu).
$$
For other metrics, the Lipschitz constant will matter. For example, take two densities $\mu,\nu$ on the real line and put $T(x)=a x$ for some $a>0$. Then
$$ d_p(\mu',\nu'):=(\int |\mu'-\nu'|^p)^{1/p} = a^{(1-p)/p}d_p(\mu,\nu).
$$
The last example is taken from Ch. 1 of https://www.szit.bme.hu/~gyorfi/nonpar_dens_en.html
; you may find the discussion there useful.