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Material Type: Notes; Class: Artificial Intelligence; Subject: Computer Science and Engineering; University: Arizona State University - Tempe; Term: Unknown 1989;
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CSE571 Artificial Intelligence Nov 12th^ Notes William Cushing
Sum over P6,P7 of P(Q,P6,P7|P4,P5) How do we do: P(Q,P6,P7|P4,P5)? = P(Q|P6,P7,P4,P5) P(P6,P7|P4,P5) = P(Q|P6,P7)P(P6,P7|P4,P5) = P(Q|P6,P7)P(P6|P7,P4,P5)P(P7|P4,P5) But, better is to do: = P(Q|P6,P7) P(P6|P4,P5) P(P7|P4,P5) Because P6 and P7 are separable (due to Q) However, that is a deeper result that depends on the structure of P6 and P An example involving a sprinkler, rain, and wet grass – the sprinkler and rain are independent, but both give wet grass. If the wet grass is unknown, then we can use the independence of the sprinkler and rain. If, however, wet grass is known, then that can’t be used. Regardless of the value that is known – true or false. For the other two relations between 3 variables, (that remain trees), the opposite is true – given information on the ‘third’ node, the first two become independent. In the above, given information, they became dependent. The generalization is known as de-separation: Given Vi and Vj (nodes of a Bayesian network) The question is: Are the two nodes conditionally independent given a set of nodes E? If, for all undirected paths between Vi and Vj There is a node X such that:
P(Q) P(P11|Q) P(P14|Q) P(P12,P13|Q) % P(P11,P12,P13,P14) No more independence, so insert parents. = P(Q) P(P11|Q) P(P14|Q) sum over P9 with P(P12,P13,P9|Q) % P(P11,P12,P13,P14) Just considering the summation P(P12,P13,P9|Q) = P(P12,P13|P9,Q) P(P9|Q) = P(P12,P13|P9) P(P9|Q) Going back... considering how to evaluate P(Q) = sum over {P6, P7} of P(Q,P6,P7) If we are evaluating one term of the sum, then we can write: P(Q|P6,P7) P(P6,P7) Where we can repeat the process to evaluate P(P6,P7) – P(Q|P6,P7) can be evaluated from the table. Also, parents introduced in this fashion are always independent, because of their child, so the above can be simplified to P(P6) P(P7) Which is one step closer to the beginning of the tree – so recursion will work.
So, all remaining are in evidence above form – which can be evaluated by recursively introducing parents. Lastly, some comment about the homework being due next week – Wednesday, I think. Should be on the web page.