Spring 2016 Section 7 (Bayes Nets + Variable Elimination) Solutions

[@zaman866]

Thanks for video. I am just wondering how we normalize to sum to 1 in part g. Can you give any numerical example? Thanks

[@hansen1101]

+Zs Sj assume f5 gives you a vector with 2 entries for +y and -y, say [1/5, 3/5]. to normalize this vector simply divide each coordinate by the sum of all coordinates [1/5 * 5/4 , 3/5 * 5/4] = [1/4, 3/4]

[@zaman866]

Thanks

[@zaman866]

hansen1101 do you know why we should normalize this and how this became non-normalized one?

[@hansen1101]

+Zs Sj in this particular case you are calculating a distribution of the form P(Q|e) where e is an instantiation of some evidence variables. By definition this form has to sum to 1 over all instances of the query variable Q (i.e. P(q1|e) + P(q2|e) = 1 in the binary case). Be careful, there are queries of other forms that need not sum to 1 and therefore normalization is not necessary (i.e P(Q,e) or P(e|Q)). This became non normalized after applying Baye's rule and only working with the term in the numerator, leavin out the joint prob. over the instantiated evidence vars in the denominator. Therefore you'll have to rescale in the end.

[@hansen1101]

concerning ex. 2f: isn't the largest factor generated 2^4? because the join on all factors containing T generates a table over 4 variables (say f2') of which one is summed out to get f2. so f2' has size 2^4

[@user-ze4qq8mm1q]

this is a good thought, but the given observation value of +z is a constant, not a variable, so although it is contained in f2(U, V, W, +z) the only variables of f2 are U, V, W, hence 2^3 = 8 .