Coarsening the conditioning variable can increase conditional mutual information

For a finite value set S, define $d(s,t)=0$ for $s=t$ and 1 otherwise; distinct endpoints force at least one intermediate distance to be 1, so the triangle inequality holds. Tolerance $1\mathrm{/}2$ makes every Cauchy sequence eventually constant, hence the metric is complete; S is itself finite and dense, and radius-$1\mathrm{/}2$ balls are singletons, so the topology is discrete. By the definitions in [F2], each such space is standard Borel; its finite products are standard Borel by the finite-product supplier. Given $Z=(i,j)$, the pair $(X,Y)$ is the point $(i,j)$, and its two conditional marginals are point masses at i and j. Their product is the point mass at $(i,j)$, so it agrees with the conditional joint law at every Z-value of positive probability. The conditional-product reference law therefore equals the joint law of $(Z,X,Y)$. By [F1], its KL divergence, and hence $I(X;Y\mid Z)$, is zero.

The variable W has one value. The joint law of $(W,X,Y)$ has mass $1\mathrm{/}2$ at $(w,0,0)$ and $1\mathrm{/}2$ at $(w,1,1)$. The conditional marginals given W are both uniform on $\{0,1\}$; their product reference law assigns mass $1\mathrm{/}4$ to each of the four pairs at w. Therefore [F1]–[F3] give $I(X;Y\mid W)=\frac{1}{2}\ln \frac{1\mathrm{/}2}{1\mathrm{/}4}+\frac{1}{2}\ln \frac{1\mathrm{/}2}{1\mathrm{/}4}=\ln 2.$

The map from Z to W is constant and thus coarsens the conditioning information. The two computed values satisfy $I(X;Y\mid W)>I(X;Y\mid Z)$, so this example rules out a general data-processing claim that coarsening the conditioning variable decreases conditional mutual information.