Conditional Independence and Mutual information I have a question concerning conditional independence. According to wikipedia (yes, maybe not the best source) two random variables are conditionally independent given a third if 
$$p(x,y|z) = p(x|z)p(y|z)\:\:\forall z.$$
However, I read a mutual information definition of conditional independence that said that two random variables are conditionally independent if 
$$I[X|Z:Y|Z] = \left\langle\left\langle\log \frac{p(x,y|z)}{p(x|z)p(y|z)}\right\rangle_{X,Y|Z}\right\rangle_Z = 0.$$
Now, I am not sure whether the two equations are equivalent. The first definitely implies the last. However, the last does not imply the first (or does it?), since there could be $z$ for which $p(x,y|z) \not= p(x|z)p(y|z)$ but the set of all $z$ for which that is true has measure zero. 
I guess the last equation means conditionally independence only almost surely (in terms of $z$) but not pointwise, but the first requires pointwise equality. 
Am I correct? If not, where is my mistake?
 A: The definition of conditional mutual information should read, in your notations,
$$I[X|Z:Y|Z] = \left\langle\left\langle\log \frac{p(x,y|z)}{p(x|z)p(y|z)}\right\rangle_{X,Y\mid Z}\right\rangle_Z,
$$
or, equivalently,
$$I[X|Z:Y|Z] =\left\langle\log \frac{p(x,y|z)}{p(x|z)p(y|z)}\right\rangle_{X,Y,Z}.$$
For discrete random variables, this is
$$I[X|Z:Y|Z] = \sum_zp(z)\sum_{x,y}p(x,y\mid z)\log \frac{p(x,y|z)}{p(x|z)p(y|z)},
$$
or, equivalently,
$$I[X|Z:Y|Z] =\sum_{x,y,z}p(x,y,z)\log \frac{p(x,y|z)}{p(x|z)p(y|z)}.$$
Then the same argument than in the unconditional case applies, namely,
$$
I[X|Z:Y|Z] = -\sum_{x,y,z}p(x,y,z)\log \frac{p(x|z)p(y|z)}{p(x,y|z)},
$$
and, since $(-\log)$ is convex,
$$I[X|Z:Y|Z] \geqslant -\log S,$$ where
$$
S=\sum_{x,y,z}p(x,y,z)\frac{p(x|z)p(y|z)}{p(x,y|z)}=\sum_zp(z)\sum_xp(x\mid z)\sum_yp(y\mid z)=\sum_zp(z)\cdot1\cdot1=1,
$$
hence $\log S=0$ and $I[X|Z:Y|Z] \geqslant0$. Furthermore, $I[X|Z:Y|Z] =0$ if and only if the convexity inequality above is an equality, that is, the function to be integrated must be constant, that is, here,
$$
\frac{p(x|z)p(y|z)}{p(x,y|z)}=c.
$$
Summing over $(x,y)$ the identities $p(x|z)p(y|z)=cp(x,y|z)$, one sees that this is only possible with $c=1$ hence $\log S=0$. Finally, $I[X|Z:Y|Z]=0$ if and only if, for every $(x,y,z)$,
$$
p(x,y|z)=p(x|z)p(y|z).
$$
