Conditional independence: does $(X \bot Y \mid Z) \land (X \bot Y \mid W) \implies (X \bot Y \mid Z , W)$? I'm reading a book about probabilistic graphical models by Daphie Koller and Nir Friedman and I'm stuck at the following exercise:

Is it true that $ (X \bot Y \mid Z) \land (X \bot Y \mid W) \implies (X \bot Y \mid Z , W) $?

Any ideas how to prove or disprove the statement? $X \bot Y \mid Z$ denotes conditional independence of X and Y given Z, i.e., $P(X\mid Z)=P(X\mid Y,Z)$.
 A: It's not true.  For a counterexample, let $X$, $Y$, and $Z$ be i.i.d., each with equal probabilities of being $+1$ and $-1$, and let $W=XYZ$.  Any three of $X$, $Y$, $Z$ and $W$ are i.i.d., but once $Z$ and $W$ are known $X$ and $Y$ are not independent as $XY=ZW$.
A: I'm learning probabilistic graphical models recently, so I'll solve this in a PGM way.
Consider a Bayesian network of the form
$X\rightarrow Z \leftarrow D \rightarrow W \leftarrow Y$
Since there are two v-structures here, the path from $X$ to $Y$ is active only when both $Z$ and $W$ are given. When only $Z$ or $W$ is given, this path is not active.
To give the above example an intuitive explanation, let's suppose $X$ and $Y$ represent the IQ of student Xavier and Ygritte, $D$ represents the difficulty of a class they have taken, $Z$ and $W$ are the scores they get in this class. If we tell you Xavier get a low score, we don't know the how that affects Ygritte's IQ. However, if both of them has a low score, things change. If we now know that Xavier has high IQ, then the probability of the class being difficult increases, even though Ygritte has a low score, that doesn't mean she's absolutely stupid, so the probability of Ygritte has high IQ increases slightly due to Xavier has high IQ.
