# sol10 - CS 70 Fall 2006 — Solutions to Homework#10 1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CS 70 Fall 2006 — Solutions to Homework #10 November 15, 2006 1. Independence. (a) Consider first any integer-valued 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 factor-set S z —we can re-write 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: • Ω = { ( i, j ) | i, j ∈ { 1 , 2 , 3 , 4 , 5 , 6 }} and so | Ω | = 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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

sol10 - CS 70 Fall 2006 — Solutions to Homework#10 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online