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While I was reading about Bayesian networks, I run into "Markov blanket" term and got severely confused with its independency in a Bayesian network graph.

Markov blanket briefly says that every node is only dependent on its parents, children and children's parents [it is gray area for node A in the picture].

Markov blanket

What is the joint probability of this BN, $P(M,S,G,I,B,R)$?

alt text
(source: aiqus.com)

If I follow the step parent only independency rule, it is:

$$ P(M | S)P(S | G,I)P(I | B)P(R | B)P(G)P(B)$$

However, if I follow the Markov Blanket independency, I end up with this (notice $P(I|\mathbf{G},B)$ is different):

$$P(M | S)P(S | G,I)P(I | \mathbf{G},B)P(R | B)P(G)P(B)$$

So which is the correct joint probability of this BN?

Update: Crosslink of this question in AIQUS

and

Respective chapter and diagrams are below:

alt text http://img828.imageshack.us/img828/9783/img0103s.png

alt text http://img406.imageshack.us/img406/3788/img0104l.png

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  • $\begingroup$ The links are all broken, could you please update them? $\endgroup$ – Lerner Zhang Jan 23 at 22:37
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Your first derivation is correct!

Because we haven't observed "Starts" or "Moves", "Ignition" is independent of "Gas". What you are writing here is just the factorisation of the joint distribution, not how to compute a the probability of a specific node given a set of observations.

What the Markov Blanket says, is that all information about a random variable in a Bayesian network is contained within this set of nodes (parents, children, and parents of children). That is, if we observe ALL OF THESE variables, then our node is independent of all other nodes within the network.

For more information about dependency within a Bayesian network, look up the concept of D-separation.

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  • $\begingroup$ thabks for the answer. But have you taken a look at the wiki page I gave. It shows a conditional probability example; implying that all MB nodes are dependent on the variable. $\endgroup$ – Özgür Feb 9 '12 at 20:56
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    $\begingroup$ The wiki page appears to be correct. The Markov Blanket is a shield from the rest of the network, such that if we know the values in that 'shield', then no other variables in the network provide any additional information about A. The key here is that we are talking about what happens when we observe those values, this doesn't change the factorization of a joint given the structure of the BN. $\endgroup$ – Nick Feb 9 '12 at 23:56

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