Given a Markov chain $X \rightarrow Y \rightarrow Z$, under what condition $I(X;Y) = I(X;Z)$ A theorem (The Data Processing Inequality) states that 

if $X \rightarrow Y \rightarrow Z$, then $I(X ; Y ) \geq I ( X ; Z )$

Question: I was wondering under what conditions $I(X;Y) = I(X;Z)$?

The proof:
Using chain rule of mutual information, we have
\begin{align*}
I ( X ; Y , Z ) &= I ( X ; Y ) + I ( X ; Z | Y )\\
&= I ( X ; Z ) + I ( X ; Y | Z )
\end{align*}
rewrite the above equalities, we have
\begin{align*}
I ( X ; Y ) + I ( X ; Z | Y ) &= I ( X ; Z ) + I ( X ; Y | Z )\\
I ( X ; Y ) &= I ( X ; Z ) + I ( X ; Y | Z )\\
I ( X ; Y ) &≥ I ( X ; Z )
\end{align*}
To obtain $I(X;Y) = I(X;Z)$ indicates $I(X;Y|Z)=0$,
\begin{align*}
 I(X;Y|Z) &= 0\\
D_{\mathrm{KL}}[p(X,Y,Z) \| p(X|Z) p(Y|Z) p(Z)] &= 0\\
p(X,Y,Z) &= p(X|Z) p(Y|Z) p(Z)
\end{align*}
Would it be possible to further simplify it for the conditions?

As stated in Cover and Thomas's Elements of Information theory 2e (in the discussion of Theorem 2.8.1, the data processing inequality), you have equality iff $X \to Z \to Y$ is a Markov chain (think of why this is equivalent to $I(X;Y|Z) =0$ and the joint distribution of $X,Y$ given $Z$ under the markovian assumption I've stated and conditional independence). 
