Data processing inequalities
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Mutual information
In a Markov chain $\BX \to \BY \to \BZ$, we have
\[
\al{
\I[\BX;\BZ] \le \min\p{\I[\BX;\BY], \I[\BY;\BZ]}.
}
\]
Intuitively, this is because both the $\BX \to \BY$ and the $\BY \to \BZ$ “channels” act as “information bottlenecks”: if information is lost in either of them, it’s lost for good.
Proof
The fact that it’s a Markov chain means that the distribution of $\BZ$ stops depending on $\BX$ once we know $\BY$: for any possible assignment $\p{x,y}$ for $\p{\BX,\BY}$, we have
\[
\bat{\BZ}{\BX=x,\BY=y} = \bat{\BZ}{\BY=y},
\]
so
\[
\al{
\I\b{\BX,\BY;\BZ}
&= \E_{(x,y) \sim (\BX,\BY)}\b{\D\bff{\at{\BZ}{\BX=x,\BY=y}}{\BZ}}\\
&= \E_{y \sim \BY}\b{\D\bff{\at{\BZ}{\BY=y}}{\BZ}}\\
&= \I\b{\BY;\BZ},
}
\]
and thus
\[
\I[\BX;\BZ] \le \UB{\I\b{\BX,\BY;\BZ} = \I\b{\BY;\BZ}}_\text{information through $\BY \to \BZ$}.
\]
Similarly, reversing the chain, $\BX$ stops depending on $\BZ$ once we know $\BY$: for any acceptable $y,z$, we have
\[
\at{\BX}{\BY=y,\BZ=z} \equiv \at{\BX}{\BY=y},
\]
so $\I[\BX;\BY,\BZ] = \I[\BX;\BY]$, and therefore
\[
\I[\BX;\BZ] \le \UB{\I[\BX;\BY,\BZ] = \I[\BX;\BY]}_\text{information through $\BX\to\BY$}.
\]
Ratio divergences
Suppose we have a “prior” variable $\BX$ and a “posterior” variable $\BX'$, and we apply the same stochastic map1 $x \mapsto \BY_x$, then the divergence of $\BY_{\BX'}$ relative to $\BY_{\BX}$ can only be smaller than the divergence of $\BX'$ relative to $\BX$:
\[
D^f\bff{\BY_{\BX'}}{\BY_\BX} \le D^f\bff{\BX'}{\BX}.
\]
That is, post-processing can only reduce the distance between distributions. Intuitively, this is because whatever noise is applied to them in the process can only cause them to “overlap more”.
Proof
Since adding data can only increase the divergence, we have
\[
\al{
D^f\bff{\BY_{\BX'}}{\BY_\BX}
&\le D^f\bff{\BX', \BY_{\BX'}}{\BX,\BY_\BX}\\
&= \E_{(x,y) \sim \p{\BX,\BY_\BX}}\b{f\p{\f{\Pr\b{\BX'=x \and \BY_{\BX'} = y}}{\Pr\b{\BX=x \and \BY_\BX = y}}}}\\
&= \E_{(x,y) \sim \BX}\b{f\p{\f{\Pr\b{\BX'=x}}{\Pr\b{\BX=x}}}}\\
&= D^f\bff{\BX'}{\BX},
}
\]
Assuming $f$ is strictly convex, equality holds iff $\BX$ and $\BX'$ can be obtained by applying the same stochastic map to both $\BY_{\BX}$ and $\BY_{\BX'}$, or equivalently iff for any connected component $S$ of the relation “has some chance of getting mapped to the same value”
\[
x_1 \mathrel{R} x_2 \ce \p{\supp\b{\BY_{x_1}} \inter\supp\b{\BY_{x_2}} \ne \es},
\]
the conditional distributions $\at{\BX}{S}$ and $\at{\BX'}{S}$ are identical.
#to-write give the formula for the gaps in either case
-
although I guess in this case it doesn’t matter whether you consider a stochastic map or a randomly-chosen function, since we’re not seeing $\BY_\BX$ and $\BY_{\BX'}$ as jointly distributed anyway ↩