How can a random variable be independent of a member of its minimal Markov blanket?

Question

Consider the following Bayes network of random variables on some probability space:

The local Markov property asserts that any variable is independent of its non-descendants given its parents. Here, $X$ has no parents, and $Y$ is a non-descendant of $X$. Thus $X$ is independent of $Y$.

A Markov blanket of the variable $X$ is any subset of other variables that "contains all the information needed to infer $X$." The Markov boundary is the smallest such subset, i.e. the Markov blanket with "no redundant information."

For a Bayes network, the Markov boundary of $X$ consists of its parents, its children, and the other parents of its children (its "spouses"). For our $X$, this is $\{Y,Z\}$.

I am perplexed that $X$ can be independent of $Y$ and yet $Y$ is still in its Markov boundary. The implication is that $Y$ provides a critical piece of information about $X$ when trying to decouple it from the rest of the network. I.e., I cannot assume $X$ is independent of $\{U,V\}$ unless I condition on both $Z$ and $Y$. And yet, somehow, $X$ is "independent" of $Y$.

This leads me to believe that maybe $Y$ is redundant, and that the Markov boundary here is really just $Z$. But then I look at any source explaining Markov boundaries and they provide explanations like this that have the same issue:

It is always true that $A$ given its parents is independent of its spouses, because the spouses are always non-descendants. Since the Markov boundary always includes the parents and children, how could the spouses have any additional information to provide?

I am struggling to understand the role of spouses in the Markov boundary, because it leads me to the following statement which I thought was definitely invalid: $$ p(X|Y) = p(X)\ \ \ \ \text{while}\ \ \ \ p(X|Y,Z) \neq p(X|Z) \tag{1} $$

Question:

How is the above Statement (1) valid?

If it's not, then doesn't $p(X|Y,Z) = p(X|Z)$ imply $Y$ isn't in the Markov boundary?

Answer 1

Addressing your question at the end, it is in fact possible for statement (1) to be true: $X$ and $Y$ may be unconditionally independent but conditionally dependent given $Z$. In the case that $Z$ is observed, the $X \to Z \leftarrow Y$ subgraph is referred to as an "activated v-structure". For an intuitive example, let $X, Y, Z$ be indicator random variables as follows:

• $X = 1$ iff my bus was late.
• $Y = 1$ iff I overslept.
• $Z = 1$ iff I am late to work.

In general $X$ and $Y$ will be unconditionally independent: knowing whether I overslept tells you nothing about whether my bus was late, in the absence of further information.

If, however, you know that I am late to work ($Z=1$), then $X$ and $Y$ become conditionally dependent: knowing that I didn't oversleep should increase your belief that my bus was late, for example.

I encourage you to try and come up with a concrete joint distribution over $X,Y,Z$ that respects the graph structure $X \to Z \leftarrow Y$ while also encoding the conditional dependence described above. (If you get stuck, check out the example from Wikipedia's article on conditional dependence.)