Probabilistic Machine Learning An Introduction, 1e
Probabilistic Machine Learning An Introduction, 1e
Exam (elaborations)
Probabilistic Machine Learning An Introduction 1st Edition By Kevin P. Murphy (Solution Manual)
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Module
Probabilistic Machine Learning An Introduction, 1e
Institution
Probabilistic Machine Learning An Introduction, 1e
Probabilistic Machine Learning An Introduction, 1e Kevin P. Murphy (Solution Manual)
Probabilistic Machine Learning An Introduction, 1e Kevin P. Murphy (Solution Manual)
Full Solution Manual for
“Probabilistic Machine Learning: An Introduction”
Kevin Murphy
1 1 Solutions
2 Part I
Foundations
3 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.
LetX1andX2be independent binary random variables, and X3=X1X2, whereis the XOR
operator. We have p(X3jX1;X2)6=p(X3), sinceX3can be deterministically calculated from X1andX2. So
the variablesfX1;X2;X3gare not mutually independent. However, we also have p(X3jX1) =p(X3), since
withoutX2, no information can be provided to X3. SoX1?X3and similarly X2?X3. HencefX1;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 havep(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) = 06= 1=8 =p(X1)p(X2)p(X3).
2.3 Conditional independence iff joint factorizes
PRIVATE
Independency)Factorization. Let g(x;z) =p(xjz)andh(y;z) =p(yjz). IfX?YjZthen
p(x;yjz) =p(xjz)p(yjz) =g(x;z)h(y;z) (3)
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