a2_sol_v2

a2_sol_v2 - THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH2603 Probability Theory Solution to Assignment 2 (Updated Version) Problem 1 (1) By Hint (2) E [ x ] = n =1 nP ( x = n ) = n =1 n i =1 P ( x = n ) = i =1 n = i P ( x = n ) = i =1 P ( X i ) = i =0 P ( X > i ) For the continuous case, we have If E [ X ] < , X non-negative, then E [ X ] = Z 0 P ( X > x ) dx Proof: Let G ( x ) = P ( X > x ) = 1 - P ( X x ) = 1 - F ( x ) E [ X ] = R 0 xf ( x ) dx = - R 0 xdG ( x ) = - ( xG ( x ) | 0 - R 0 G ( x ) dx ) = - xG ( x ) | 0 + R 0 G ( x ) dx Claim: lim x →∞ xG ( x ) = 0 Proof: xG ( x ) Z x tf ( t ) dt lim x →∞ xG ( x ) lim x →∞ Z x tf ( t ) dt = 0 since E ( X ) < Problem 45 Let N denote the number of collisions, and let X denote the number of nonempty boxes, then N = r - X Then let I i = ± 1 if box i is nonempty 0 otherwise X = I 1 + I 2 + ... + I k E [ X ] = E [ I 1 ] + E [ I 2 ] + ... + E [ I k ] = k i =1 P (box i is nonempty) = k i =1 (1 - P (box i is empty)) = k i =1 (1 - (1 - p i ) r ) 1
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E ( N ) = r - E [ X ] = r - k + k X i (1 - p i ) r Alternatives: Let N = N 1 + N 2 + ... + N k , where N i is the number of collisions in box i . More discussion:
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This note was uploaded on 01/02/2012 for the course MATH 2603 taught by Professor Han during the Spring '10 term at HKU.

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a2_sol_v2 - THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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