Full Solution Manual for “Probabilistic Machine Learning: An Introduction”

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Full Solution Manual for “Probabilistic Machine Learning: An Introduction” 2 Solutions 2.1 Conditional independence PRIVATE 1. Bayes’ rule gives P (HjE1; E2) = P (E1; E2jH)P (H) P (E1; E2... ) (1) Thus the information in (ii) is sufficient. In fact, we don’t need P (E1; E2) because it is equal to the normalization constant (to enforce the sum to one constraint). (i) and (iii) are insufficient. 2. Now the equation simplifies to P (HjE1; E2) = P (E1jH)P (E2jH)P (H) P (E1; E2) (2) so (i) and (ii) are obviously sufficient. (iii) is also sufficient, because we can compute P (E1; E2) using normalization. 2.2 Pairwise independence does not imply mutual independence We provide two counter examples. Let X1 and X2 be independent binary random variables, and X3 = X1 ⊕ X2, where ⊕ is the XOR operator. We have p(X3jX1; X2) 6= p(X3), since X3 can be deterministically calculated from X1 and X2. So the variables fX1; X2; X3g are not mutually independent. However, we also have p(X3jX1) = p(X3), since without X2, no information can be provided to X3. So X1 ? X3 and similarly X2 ? X3. Hence fX1; X2; X3g are pairwise independent. Here is a different example. Let there be four balls in a bag, numbered 1 to 4. Suppose we draw one at random. Define 3 events as follows: • X1: ball 1 or 2 is drawn. • X2: ball 2 or 3 is drawn. • X3: ball 1 or 3 is drawn. We have p(X1) = p(X2) = p(X3) = 0:5. Also, p(X1; X2) = p(X2; X3) = p(X1; X3) = 0:25. Hence p(X1; X2) = p(X1)p(X2), and similarly for the other pairs. Hence the events are pairwise independent. However, p(X1; X2; X3) = 0 6= 1=8 = p(X1)p(X2)p(X3). 2.3 Conditional independence iff joint factorizes PRIVATE Independency ) Factorization. Let g(x; z) = p(xjz) and h(y; z) = p(yjz). If X ? Y jZ then p(x; yjz) = p(xjz)p(yjz) = g(x; z)h(y; z) (3) 4 [Show More]

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