Conditional mutual information and Markov chain. If we have the Markov chain $X \to Y \to Z$, or equivalently
$$I(X;Z| Y)=0, \tag{1}$$
where $I(\cdot)$ denotes the mutual information.
Does the Markov chain $X \to (Y,W) \to Z$ also hold? 
Or 
$$I(X;Z|Y,W)=0~~? \tag{2}$$
Intuitively, I think (2) is true. But how to prove? Thanks in advance.
Another simple question about probabilities,
$$P(Z=z|X=x,Y=y)\cdot P(Y=y)=P(Z=z,Y=y|X=x) \tag{3}$$
Is (3) right? I am confused. Thanks again.
 A: The answer to your first question is actually no, the second one Markov chain need not be correct. To see this, take a trivial case where $X$ and $Z$ are independent variables and $Y$ is a trivial variable (only one element in its range with probability $1$). Trivially $X$-$Y$-$Z$ form a Markov chain. Now take $W=X \oplus Z$ (i.e. sum modulo $2$), so that it holds the parity of $X$ and $Z$. Now, clearly, knowing $W$ correlates $X$ and $Z$ (if $W=0$, then $X=Z$; if $W=1$, then $X \ne Z$), so that $I(X:Z|W) \ne 0$ (or equivalently $I(X:Z|WY) \ne 0$, as $Y$ is a trivial random variable and including it doesn't harm anything). In other words $X$-$WY$-$Z$ is no longer a Markov chain.
Your second question is regarding the extension of the chain rule to conditional probabilities:
$$P(A_n,\dots,A_1 \mid B) = P(A_n \mid A_{n-1},\dots,A_1,B) \cdot P(A_{n-1},\dots,A_1 \mid B) \ ,$$
so the correct form is:
$$P(Z=z \mid Y=y, X=x) \cdot P(Y=y \mid X=x) = P(Z=z, Y=y \mid X=x) \ .$$
A: The first is practically trivial, once we understand that the random variables $X$, $Y$... are not necessarily one-dimensional (scalars). So you can just define $S=(Y,W)$ (a multivariate variable) and with that notation you get the first case.
Regarding the second, simplifying/abusing notation:
$$P(Z | X Y ) P(Y) = \frac{P(Z X Y)}{P(XY)} P(Y)$$
and
$$ P(Z Y |X)=\frac{P(Z X Y)}{P(X)}$$
These would be equal iff $\frac{P(Y)}{P(XY)}=\frac{1}{P(Y)}$, or $P(XY)=P(X)P(Y)$, i.e. iff $X,Y$ are independent.
