ps2-sol-07

ps2-sol-07 - CS228 Problem Set #2 Solutions 1 CS 228,...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CS228 Problem Set #2 Solutions 1 CS 228, Winter 2007 Problem Set #2 Solutions Handout #12 For each problem, a number of error codes describing common mistakes made by students are listed below. If you feel that your homework has been wrongly graded, please come see us. All error codes are tentative in this handout. They will go through modifications as we grade the homeworks. 1. [20 points] In this problem, you will show that the set of variables in a clique tree separator d-separates the graph into two conditionally independent pieces. (a) [6 points] For an undirected graph H , recall that we say sep H ( X ; Y | Z ) (i.e., X and Y are separated given evidence Z in H ) whenever all paths between variables in X and variables in Y are blocked by some variables in Z . Prove that if we have a moralized graph H derived from an original Bayesian network G then sep H ( X ; Y | Z ) implies d- sep G ( X ; Y | Z ). This is a generalization of your result from Problem 5 on the previous problem set, and will be used in the next part. Answer: Suppose for a contradiction that we have sep H ( X , Y | Z ), but not d- sep G ( X , Y | Z ). This means that in G with Z observed, there exist Z X and Z k Y with an active trail Z Z 1 Z k . For the trail to be active, it must be the case that for every i : 0 < i < k , the triplet Z i- 1 Z i Z i +1 is active. For each Z i whose triplet is not a v-structure, certainly Z i cannot be in Z , or the trail would not be active. For any Z i which sits at the base of a v-structure Z i- 1 Z i Z i +1 , we see that neither of its neighbors can also be a v-structure in the trail, due to the directions of their arrows toward Z i , and hence from above, we know that neither of them is in Z . We also know that since H is moralized, it must contain an edge Z i- 1- Z i +1 . Thus we can easily construct a path in H from Z to Z k by following the active trail of G , but shortcutting each v-structure via its moral edge. By this construction, none of the nodes along this path will be in Z , and hence the path must be active, contradicting the assumption that Z separates X from Y in H . (b) [14 points] Now, assume that we perform variable elimination on our Bayesian net- work G , using the Sum-Product-Variable-Elimination algorithm (see page 279). This defines an induced graph I where each intermediate factor during the elimination process defines a clique (see section 8.2.2). Furthermore, as described in class this defines a clique tree T where each intermediate factor is represented as a clique in the tree and therefore as a subset of a maximal clique in I . Let S be a clique separator in the clique tree T . Let X be the set of all variables mentioned in T on one side of the separator and let Y be the set of all variables mentioned on the other side of the separator. Prove that sep I ( X ; Y | S ) and con- clude (using the above result) that d- sep G ( X ; Y | S ). (Hint: Remember the running)....
View Full Document

Page1 / 10

ps2-sol-07 - CS228 Problem Set #2 Solutions 1 CS 228,...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online