Why is $ (X \perp Y\mid Z) \ \& \ (X \perp W \mid Z) \implies X \perp (Y,W) \mid Z $ false? Is there a counter example?

Question

I am learning probabilistic graphical model recently. I can't construct a graphical model that contradicts the aforementioned rule, whether Bayesian or Markov network.

If $ (X \perp Y\mid Z) $, then there's no active path between $X$ and $Y$ given $Z$, and if $(X \perp W \mid Z) $, then there's no active path between $X$ and $W$ given $Z$. So there's no active path between $X$ and $Y,W$ given $Z$, hence $X \perp (Y,W) \mid Z $. What is missing here?

Thanks in advance.

Answer 1

Let $X,Y\sim\operatorname{i.i.d. Bernoulli}(1/2)$ and let $W$ be the mod-$2$ sum of $X$ and $Y;$ thus $(X,Y) = (0,0),(0,1),(1,0),(1,1)$ each with probability $1/4$ and $$ W= \begin{cases} 0 & \text{if } X=Y, \\ 1 & \text{if } X\ne Y. \end{cases} $$ Then let $Z=X.$

Then $Y$ and $X$ are conditionally independent given $Z,$ and $X$ and $W$ are conditionally independent given $Z,$ but $X$ is not conditionally independent of $(Y,W)$ given $Z.$