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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 ] = 0 P ( X > x ) dx Proof: Let G ( x ) = P ( X > x ) = 1 - P ( X x ) = 1 - F ( x ) E [ X ] = 0 xf ( x ) dx = - 0 xdG ( x ) = - ( xG ( x ) | 0 - 0 G ( x ) dx ) = - xG ( x ) | 0 + 0 G ( x ) dx Claim: lim x →∞ xG ( x ) = 0 Proof: xG ( x ) x tf ( t ) dt lim x →∞ xG ( x ) lim x →∞ 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 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: How to show that E ( N ) = r - k + k i =1 (1 - p i ) r 0? Note that (1 - x ) r 1 - rx Then E ( N ) r - k + k i =1 (1 - p i r ) = 0 Problem 74 (a) Method 1 P (a record occurs at time n ) = P ( X n > X 1 , X n > X 2 , . . . , X n > X n - 1 ) = -∞ x n -∞ x n -∞ . . .
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