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Unformatted text preview: CS 70 Fall 2006 Solutions to Homework #10 November 15, 2006 1. Independence. (a) Consider first any integervalued random variables X and Y . By the definition of expectation: E [ XY ] = z Z z Pr ( XY = z ). Focus on one such { XY = z } . This event is equivalent to the event S a,b Z : z = ab { X = a Y = b } ; and thus since the events in the union are pairwise disjoint it follows that for each z Z : Pr ( XY = z ) = Pr [ a,b Z : z = ab X = a Y = b = X a,b Z : z = ab Pr ( X = a Y = b ) . And since the sets S z = ( a, b ) Z 2  z = ab , for z Z , partition Z 2 i.e. each z Z has the set S z of pairs of factors whose product is z and each pair of integers ( a, b ) belongs to exactly one such factorset S z we can rewrite the original sum defining E [ XY ] as follows: E [ XY ] = X z Z z Pr ( XY = z ) = X a Z X b Z ab Pr ( X = a Y = b ) . This formally proves the hint. Now given independence of X and Y we can get: E [ XY ] = X a Z ( X b Z ab Pr ( X = a Y = b ) ) = X a Z ( X b Z ab Pr ( X = a ) Pr ( Y = b ) ) = X a Z ( a Pr ( X = a ) X b Z b Pr ( Y = b ) ) = X a Z { a Pr ( X = a ) E [ Y ] } = E [ Y ] X a Z { a Pr ( X = a ) } = E [ Y ] E [ X ] . 1 (b) Consider now a dependent case: X = ( 2 with probability 0 . 5 with probability 0 . 5 Y = X . Then we can see that the decomposition of Part (a) does not hold, since E [ XY ] = E X 2 = 0Pr ( X = 0) + 4Pr ( X = 2) = 2 6 = 1 = (2 . 5) 2 = (0Pr ( X = 0) + 2Pr ( X = 2)) 2 = ( E [ X ]) 2 = E [ X ] E [ Y ] . 2. Pairwise Independence. (a) Let us first enumerate the sample space and the three events of interest: = { ( i, j )  i, j { 1 , 2 , 3 , 4 , 5 , 6 }} and so   = 36; A = { (1 , j )  j { 1 , . . . , 6 }} { (3 , j )  j { 1 , . . . , 6 }} { (5 , j )  j { 1 , . . . , 6 }} and so  A  = 18; B = { ( i, 1)  i { 1 , . . . , 6 }} { ( i, 3)  i { 1 , . . . , 6 }} { ( i, 5)  i { 1 , . . . , 6 }} and so  B  = 18; and C = { ( i, j )  i, j { 1 , . . . , 6 } , i even, j odd }{ ( i, j )  i, j { 1 , . . . , 6 } , i odd, j even } and so  C  = 18....
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 Fall '08
 HAULLGREN

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