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The following Bayesian network contains a node which is deterministically dependent on its parents: the variable $either$ is simply the $OR$ function of its parents $tub$ and $lung$.

By the graph, the Markov blanket of $either$ is the set $\{tub, ~lung, ~xray, ~dysp, ~bronc\}$, namely the parents, children and spouses of $either$. But $either$ is a function of $tub$ and $lung$, completely determined by them. It doesn't add any extra information (except being $OR$). It is completely independent of its children given its parents, let alone the entire network. Thus, can the Markov blanket be just $tub$ and $lung$?

I implemented a Markov blanket finding algorithm myself (IPC-MB), and it returns only $\{tub, lung\}$ as the Markov blanket, using the G-test as a conditional independence test. But by the d-separation criterion applied on the graph of the network, that wouldn't be enough.

What is the actual Markov blanket of $either$, knowing it is a function of its parents?

The Bayesian network "ASIA" from bnlearn

(taken from the Bayesian network repository at bnlearn)

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    $\begingroup$ The Markov blanket of a node in a Bayesian network consists of the set of parents, children and spouses (parents of children), under certain assumptions. One of them is the faithfulness assumption, which, together with the Markov condition, implies that two variables X and Y are conditionally independent given a set of variables Z if and only if they are d-separated by Z in the graph. Deterministic relationships violate the faithfulness assumption. For example, in your case "either" is conditionally independent of all variables conditional on {"tub", "lung"}, although they are not d-separated. $\endgroup$ – George Jun 29 at 10:21
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    $\begingroup$ @George, your comment is the actual answer to my question. Can you please post it as an answer, so I can accept it? $\endgroup$ – CamilB Jun 29 at 12:27
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The Markov blanket of a node in a Bayesian network consists of the set of parents, children and spouses (parents of children), under certain assumptions.

One of them is the faithfulness assumption, which, together with the Markov condition, implies that two variables $X$ and $Y$ are conditionally independent given a set of variables $Z$ if and only if they are d-separated by $Z$ in the graph. Deterministic relationships violate the faithfulness assumption.

For example, in your case $either$ is conditionally independent of all variables conditional on $\{tub, lung\}$, although they are not d-separated.

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  • $\begingroup$ This clears it up, thank you. The graph in my question could be fixed by making $xray$ and $dysp$ as children of both $tub$ and $lung$, instead of being children of $either$, right? In that case, $either$ would end up with no children, thus it would indeed be d-separated from the entire network by its parents. $\endgroup$ – CamilB Jun 30 at 13:19

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