If $X \to Y \to Z$ is a Markov chain, prove or disprove $H(Y\mid X)\le H(Z\mid X)$.
I said the statement was true, and from $I(X;Y)\ge I(X;Z)$ by definition, thus $H(X) - H(X\mid Y) \ge H(X)-H(X\mid Z)$, but it's clear this was wrong since I used a different inequality. How should I have proven/disproven this statement?
The straighforward inequality is $$H(Z|Y)=H(Z|Y,Z)\le H(Z|X) \tag{1}$$ (first equality comes from Markov property; second inequality comes from chain rule).
It doesn't seem that this can be linked with $$H(Y∣X) \stackrel{?}{\le} H(Z∣X) \tag{2}$$
So, let's look for a counterexample. Here's a trivial one: let $X,Y$ be iid Bernoulli, and $Z$ be a constant. This is a (rather silly, but valid) Markov chain: $P(Z|Y,X)=P(Z|Y)$ But $H(Z| X) =H(Z)=0$ and $H(Y|X)=H(Y)=1$