Let $P$ be a probability distribution and let $A$ and $B$ be some events, and suppose that we want to minimise an $f$-divergence between $P$ and the set of all distributions $Q$ that satisfy that constraint that $Q(B|A) = q$ for some fixed $q \geq P(B|A)$. Let $P\_{f}$ denote the result of minimising a given $f$-divergence with this constraint. Is it always true, for any $f$, that $P\_{f}(A) \leq P(A)$? I know that this holds for several examples (Kullback Leibler divergence, Hellinger distance, Inverse Kullback Leibler divergence), but is it true in general? 



Minimising the f-divergence to a conditional probability constraint
After having some trouble understanding this question, I have come to interpret it as follows: Let $P$ be a probability measure on a measurable space $(S,\Sigma)$. Suppose that $A$ and $B$ are in $\Sigma$, and $P(A)>0$. Fix any $q\in[P(B|A),1]$. Let $f\colon[0,\infty]\to(-\infty,\infty]$ be a function that is convex and continuous on $[0,\infty]$ and finite on $(0,\infty)$. Let $Q\_f$ be a minimizer of the $f$-divergence 
$$D\_f:=D\_f(P||Q):=\int\_S f\Big(\frac{dP}{dQ}\Big)\,dQ$$
over the set of all probability measures $Q$ on $(S,\Sigma)$ such that $Q(B|A)=q$. Is it then necessarily true that $Q\_f(A)\le P(A)$? 


The answer to this question is no. Indeed, let $S=\{1,2,3\}$, $A=\{1,2\}$, $B=\{2,3\}$, $P(\{i\})=p\_i$ where $(p\_1,p\_2,p\_3)=(1,1,2)/4$, $q=4/5$, and $f(x)=(4-3x)\_++(1-2x)\_+$, where $u\_+:=\max(0,u)$. 


Introducing now $k\_i:=Q(\{i\})/P(\{i\})$, rewrite the conditions that $Q(B|A)=q$ and that $Q$ is a probability measure as $\frac{k\_2}{k\_1+k\_2}=\frac45$ and $\frac14\,(k\_1+k\_2)+\frac12\,k\_3=1$, which can be further rewritten as $k\_1=\frac45\,(1-k\_3/2)$ and $k\_2=\frac{16}5\,(1-k\_3/2)$, with the restriction $0\le k\_3\le2$ to ensure that $k\_1,k\_2,k\_3$ are all nonnegative. So, we can write 
\begin{equation}
 D\_f(P||Q)=F(k\_3):=\sum\_1^3 f(1/k\_i)k\_i p\_i
 =
 \begin{cases}
 \frac{2}{5} (7-6 k\_3) & \text{if } k\_3\le\tfrac18, \\
 \frac{11}{4}-2 k\_3 & \text{if } \frac{1}{8}\leq k\_3\leq \frac{3}{4} \\
 \frac{1}{20} (8 k\_3+19) & \text{if } \frac{3}{4}\le k\_3\le\frac{49}{32}, \\ 
 2 k\_3-\frac{3}{2} & \text{if } \frac{49}{32}\le k\_3. 
\end{cases}
\end{equation}
Thus, the minimization of $D\_f(P||Q)$ in $Q$ such that $Q(B|A)=q$ here reduces to the minimization of the (convex) function $F$ on the interval $[0,2]$. It is obvious that the unique minimizer here is $k\_3=\frac34$, which then leads to $k\_1=\frac12$, $k\_2=2$, and $Q\_f(A)=k\_1p\_1+k\_2p\_2=\frac18+\frac12>\frac12=P(A)$. So, the inequality $Q\_f(A)\le P(A)$ fails to hold. 




---


**Remark.** It may be of interest how the above counterexample was obtained; that was based on a few not so straightforward observations. The difficulty here is that the convex function $f$ and the measures $P$ and $Q$ are, in general, infinite-dimensional objects. We minimize $D\_f(P||Q)$ in $Q$ subject to the restriction $Q(B|A)=q$, which can be rewritten as $Q(A\cap B)=qQ(A)$. Note that this restriction is (i) linear in $Q$ and (ii) depends on $Q$ only through its restriction to the small sigma algebra $\Sigma\_0$ generated by the partition $(A\setminus B,A\cap B,S\setminus A)$ of $S$. 


On the other hand, $D\_f(P||Q)=\int\_S g(K)\,dP$, where $K:=\frac{dQ}{dP}$ and 
\begin{equation\*}
 g(k):=k\,f(\tfrac1k),\quad\text{with }g(0):=g(0+),\ g(\infty):=g(\infty-). \tag{1}
\end{equation\*}
The latter formula defines a map of the set of all functions $f\colon[0,\infty]\to(-\infty,\infty]$ that are convex and continuous on $[0,\infty]$ and finite on $(0,\infty)$ onto the same set (in fact, this map is an involution and hence bijective). So, by Jensen's inequality, 
\begin{equation\*}
 D\_f(P||Q)=E\_P g(K)=E\_P\,E\_P(g(K)|\Sigma\_0)\ge E\_P g(K\_0),\quad\text{where } K\_0:=E\_P(K|\Sigma\_0) 
\end{equation\*}
and $E\_P$ denotes the expectation with respect to the measure $P$. So, without loss of generality (wlog) $K$ is $\Sigma\_0$-measurable and hence 
takes constant values, say $k\_1,k\_2,k\_3$, on the sets $A\setminus B,A\cap B,S\setminus A$. That is, wlog $S=\{1,2,3\}$, as in the above counterexample. 
Thus, the measures $P$ and $Q$ are no longer infinite-dimensional objects; they are now completely represented by their values on the sets $A\setminus B,A\cap B,S\setminus A$: say $p\_1,p\_2,p\_3$ for $P$ and hence $k\_1p\_1,k\_2p\_2,k\_3p\_3$ for $Q$. Accordingly, now we can write 
\begin{equation\*}
 D\_f(P||Q)=\sum\_1^3 g(k\_i)p\_i. 
\end{equation\*}
We need to minimize this in $k\_1,k\_2,k\_3\ge0$ subject to the linear restrictions $k\_1p\_1+k\_2p\_2+k\_3p\_3=1$ and $k\_2p\_2=q(k\_1p\_1+k\_2p\_2)$, the latter restriction representing $Q(A\cap B)=qQ(A)$. 
The condition $q \geq P(B|A)$ can now be rewritten as $vp\_2=q(p\_1+p\_2)$ for some $v\ge1$. 
Writing the Lagrange multipliers system, we see that the dependence on the convex function $g$, an infinite-dimensional object, is reduced to the dependence on just the three numbers $h\_i:=g'(k\_i)$, such that the function $k\_i\mapsto h\_i$ is nondecreasing, that is, $(h\_i-h\_j)(k\_i-k\_j)\ge0$ for all $i,j$ in $\{1,2,3\}$. 


The inequality $Q\_f(A)\le P(A)$ in question in the (now justified) reduced setting can be rewritten as $k\_1p\_1+k\_2p\_2\le p\_1+p\_2$. Thus, it remains to check whether the latter inequality is implied by the system of algebraic equations and inequalities described above, involving the 10 real variables $p\_1,p\_2,p\_3,k\_1,k\_2,k\_3,v,h\_1,h\_2,h\_3$. With the help of a computer algebra package, we find the above counterexample showing that the implication in question fails to hold in general. (Given the numbers $h\_i$ such that $(h\_i-h\_j)(k\_i-k\_j)\ge0$ for all $i,j$, we can find any number of convex functions $g$ such that $h\_i=g'(k\_i)$ for $i=1,2,3$; I took one of such functions $g$ and then obtained $f$ from $g$ using the involution formula (1).) 


