hw1sol - 10708 Graphical Models: Homework 1 Solutions...

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10708 Graphical Models: Homework 1 Solutions October 13, 2006 1 Conditional Probability 1.1 Prove P ( S ) = f ( X,Z ) g ( Y,Z ) = ( X Y | Z ) P ( X,Y | Z ) = P ( X,Y,Z ) P ( Z ) = P ( X,Y,Z ) x y P ( x,y,Z ) = f ( X,Z ) g ( Y,Z ) x f ( x,Z ) y g ( y,Z ) = P ( X | Z ) P ( Y | Z ) Prove ( X Y | Z ) = P ( S ) = f ( X,Z ) g ( Y,Z ) P ( S ) = P ( X,Y,Z ) = P ( X,Y | Z ) P ( Z ) = P ( X | Z ) P ( Y | Z ) P ( Z ) where f ( X,Z ) = P ( X | Z ) and f ( Y,Z ) = P ( Y | Z ) P ( Z ). 1.2 Assume that the joint can be expressed as the product of marginals, P ( X,Y,Z ) = f ( X,Z ) g ( Y,Z ). Note that P ( Z ) = x f ( x,Z ) = y g ( y,Z ). It follows that X x X y P ( x,y,Z ) = X x X y f ( x,Z ) g ( y,Z ) P ( Z ) = P ( Z ) P ( Z ) which is only true is P ( Z = z ) = 1 for some z . For general distributions, the statement does not hold. 1
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1.3 Prove ( X Y,W | Z ) = ( X Y | Z ) P ( X,Y | Z ) = X w P ( X,Y,w | Z ) = X w P ( X | Z ) P ( Y,w | Z ) = P ( X | Z ) P ( Y | Z ) Prove ( X Y | Z ) ( X,Y W | Z ) = ( X W | Z ) P ( X,W | Z ) = X y P ( X,y,W | Z ) = X y P ( X,y | Z ) P ( W | Z ) = X y P ( X | Z ) P ( y | Z ) p ( W | Z ) = P ( X | Z ) P ( W | Z ) Prove ( X Y,W | Z ) ( Y W | Z ) = ( X,W Y | Z ) P ( X,W,Y | Z ) = P ( X | Z ) P ( Y,W | Z ) = P ( X | Z ) P ( Y | Z ) P ( W | Z ) (i) P ( X,W | Z ) = X y P ( X,W,y ) = P ( X | Z ) P ( W,y | Z ) = X y P ( X | Z ) P ( W | Z ) P ( y | Z ) = P ( X | Z ) P ( W | Z ) (ii) combining equations (i) and (ii) gives us that P ( X,W,Y | Z ) = P ( X | Z ) P ( X,W | Z ). 1.4 The simplest example uses exclusive-or ( ). Let X 1 ,X 2 Bernoulli( 1 2 ) with X 3 = X 1 X 2 . It is easy to show that P ( X i ,X j ) = 1 4 = P ( X i ) P ( X j ). Since X 3 depends deterministically on X 1 and X 2 it cannot be the case that ( X 1 ,X 2 X 3 ). 2 Graph Independencies 2.1 The minimal subset of the variables A = { X 2 ,X 3 ,X 4 ,X 5 ,X 8 } . X 2 and X 3 must be in the subset because they are children of X 1 . X 4 and X 8 must be in the subset because they are the parents of X 1 . X 5 must be in the subset because including X 3 creates an active trail from X 1 to X 5 . Given this subset A , the rest of the variables are d-separated from X 1 . 2
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2.2 The minimal subset of the variables is the parents of X i , the children of X i , and the parents of the children of X i . Necessity: The parents of
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This note was uploaded on 05/25/2008 for the course MACHINE LE 10708 taught by Professor Carlosgustin during the Fall '07 term at Carnegie Mellon.

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hw1sol - 10708 Graphical Models: Homework 1 Solutions...

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