Let $P$ and $Q$ be two distributions over a sample space $\Omega$ which I would like to show are close under some choice of distance function. So far I have managed to show that there exists a subset $S\subseteq \Omega$ such that:


* $P(S)$ is large, say, at least $(1-\varepsilon)$
* the conditional distributions of $P$ and $Q$ over $S$, denoted $P\_S(x)=P(x)/P(S)$ and $Q\_S(x)=Q(x)/Q(S)$ are pointwise close, i.e. $P\_S(x)\in [1\pm\varepsilon]\;Q\_S(x)$ for any $x\in S$.


My question is: does this correspond to a distributional closeness between $P$ and $Q$ under any well-known divergence? I realize that the above notion of closeness is asymmetric as the first point above is with respect to one of the distributions.



What is this distributional closeness?
$\newcommand\Om\Omega$No. E.g., for natural $n$, suppose that $\Om=[n]:=\{1,\dots,n\}$, $S=[n-1]$, $P(x)=\frac1n$ for $x\in\Om$, $Q(x)=\frac1{n^2}$ for $x\in S$, and $Q(n)=1-\frac{n-1}{n^2}$.


Then your conditions hold for $\varepsilon=\frac1n$, $P$ is uniform over $\Om$, but (for large $n$) almost all $Q$-mass is at the one point, $n$.




---


A probability distribution is a measure. So, you should write $P(\{x\})$ instead of $P(x)$, assuming that the singleton sets $\{x\}$ are in the underlying $\sigma$-algebra -- which, looking at the context of your post, appears to be the largest $\sigma$-algebra over a discrete set $\Om$. However, in the answer above I used your notations, such as $P(x)$. Also, $P\_S(x)$ is a number, not a distribution, conditional or not.


