Consider a distribution $P(X,Y,Z)$ and a Markov chain $Z-Z'$. Does the following equality hold in general? $$ I(X;Y \mid Z) = I(X;Y \mid Z,Z') $$

  • 1
    $\begingroup$ What's a Markov chain $Z-Z'$? Do you mean that $Z'$ is the next state in the chain and $Z$ is the previous state? $\endgroup$ Jan 18, 2021 at 10:15
  • $\begingroup$ I mean that $Z'$ is derived from $Z$ through a (deterministic or random) function. $\endgroup$
    – Cesare
    Jan 18, 2021 at 10:17
  • $\begingroup$ Ok this makes sense $\endgroup$ Jan 18, 2021 at 10:18

1 Answer 1


Yes, it does hold in general. You can't create new information by performing deterministic or random operations on a variable, only destroy information or keep information the same. So all of the info in $Z'$ is already contained in $Z$. See wiki for proof for entropies. Proof for conditional mutual information follows analogously.

Edit: Ok, here's a formal derivation. Firstly, let's formalize the Markov Chain by saying that $Z' = f(Z, G)$, where $f$ is some deterministic function, and $G$ is some random variable (possibly vector-valued), which is completely unrelated to $Z$ and which denotes the random part of the step function. Now let's try to find the joint entropy $H(Z, Z')$

$$H(Z, Z') = H(Z) + H(Z' | Z) = H(Z) + K_1$$

This is a sum of two numbers. The first number only depends on the probability distribution of $Z$. The second number (which I will call K_1) explicitly does not depend on the distribution of $Z$, because we have conditioned it out. Hence, it will only depend on $G$. How exactly it depends on $G$ turns out to be irrelevant. We can also extend this result by assuming there is one or more extra variables in the expression

$$H(X, Y, Z, Z') = H(X, Y, Z) + H(Z' | X, Y, Z) = H(X, Y, Z) + K_2$$

Next, we will transform the original expression and try to apply this result

$$I(X;Y| Z, Z') = H(X,Y,Z,Z') - H(Z,Z') = H(X, Y, Z) + K_2 - H(Z) - K_1 = I(X; Y | Z) + K_2 - K_1$$

So what remains to be proven is that $K_1 = K_2$, or, explicitly

$$H(Z'|Z) = H(Z'|X,Y,Z)$$

Again, this last one seems completely obvious logically, namely, that conditioning using variables that do not deliver extra information compared to what is already conditioned upon does not change anything. I can't immediately recall how to prove this last one. I'll look it up when I get the chance

  • $\begingroup$ Do you have a formal derivation? $\endgroup$
    – Cesare
    Jan 18, 2021 at 11:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.