Let $X,Y,Z,Y',Z'$ be random variables where $Y\sim Y', Z\sim Z'$, $Y$ and $Z$ are independent, while $Y'$ and $Z'$ are, in the sense that we have
- $p(X,Y,Z)=p(X|Y,Z)p(Y)p(Z)$
- $p(X,Y',Z')=p(X|Y',Z')p(Y',Z')$
- $p(X|Y,Z)=p(X|Y',Z')$
Is $H(X|Y,Z)=H(X|Y',Z')$?
It seems whether $p(Y,Z)$ factorises or not does not affect the value of $H(X|Y,Z)$ but somehow the rigorous argument is missing.
The statement is not clear, because you don't specify how $Z$ is (jointly) related with the other variables. Assuming you mean that the conditional are the same ($P(X|Y,Z)=P(X|Y',Z')$) : $$H(X|Y,Z)=-\sum_{y,z} p(y,z) \sum_{x} p(x|y,z)\log p(x|y,z) $$ This implies that for both scenarios the inner sum (the entropies conditioned on particular values: $H(X|Y=y,Z=z)$) are the same. Hence, in general the conditined entropy will not depend solely on the marginals. That is, $H(X|Y,Z) \ne H(X|Y',Z') $