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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 impossible: $$ p(X|Y) = p(X)\ \ \ \ \text{while}\ \ \ \ p(X|Y,Z) \neq p(X|Z) \tag{1} $$

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?

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?

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 $X$. Therefore $X$ 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 $Y$.

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 impossible? $$ p(X|Y) = p(X)\ \ \ \ \text{while}\ \ \ \ p(X|Y,Z) \neq p(X|Z) \tag{1} $$

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?

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 impossible: $$ p(X|Y) = p(X)\ \ \ \ \text{while}\ \ \ \ p(X|Y,Z) \neq p(X|Z) \tag{1} $$

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?

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 $X$. Therefore $X$ 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 $Y$.

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 impossible: $$ p(X|Y) = p(X)\ \ \ \ \text{while}\ \ \ \ p(X|Y,Z) \neq p(X|Z) \tag{1} $$

Question:

Is the above equation possible? If not, then how is $X$ independent of $Y$ while $Y$ is in the (minimal) Markov-boundary of $X$? If it's not, then how is $Y$ different from the spouses in the other diagram?

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 $X$. Therefore $X$ 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 $Y$.

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 impossible? $$ p(X|Y) = p(X)\ \ \ \ \text{while}\ \ \ \ p(X|Y,Z) \neq p(X|Z) \tag{1} $$

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?