When can conditional mutual information be decomposed as a sum?

Question

More specifically: What are the necessary conditions to be able to write the following? $$I(X;Y|Z) = \sum_z p(z) \cdot I(X;Y|Z=z)$$

Isn't this always possible, since I can always write $p(x,y,z) = p(z) \cdot p(x,y | z)$? Or is there some independence condition I don't see?

Answer 1

It always holds. Proof (credit goes to my friend):

\begin{align} I(X;Y|Z) &= H(X|Z) + H(Y|Z) - H(X,Y|Z)\\ &= \sum_z p(z) (H(X|Z = z) + H(Y|Z = z) - H(X,Y|Z = z)\\ &= \sum_z p(z) I(X;Y | Z=z) \end{align}