STAT_333_Assignment_2_Solutions

STAT_333_Assignment_2_Solutions - STAT 333 Assignment 2...

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Unformatted text preview: STAT 333 Assignment 2 SOLUTIONS Due: Thursday, June 26 at the beginning of class 1. Suppose we toss a fair coin repeatedly and observe a sequence of H or T. Let λ be the event “H H T T ”. a. Why is λ a renewal event? There is no overlap, so knowing that the event has just occurred doesn’t help us get it any earlier. Alternatively, the waiting time for the event to occur the first time is the same as if it has just occurred. b. Use the renewal sequence { r n } to show that λ is recurrent. r = 1, r 1 = r 2 = r 3 = 0, r 4 = (½) 4 , r n = (½) 4 , n ≥ 4. E[V λ ] = ∑ n =1 to ∞ r n = ∑ n =4 to ∞ (½) 4 = ∞ so λ is recurrent. c. Determine the generating function of the renewal sequence, ( ) R s λ , use it to obtain ( ) F s λ and prove recurrence with this method. , 1 16 1 ) ( 1 ) ( 4 4 4 2 1 s s s s r s R n n n n n- + = + = = ∑ ∑ ∞ = ∞ = λ for | s | < 1 16 1 16 16 1 1 1 ) ( 1 1 ) ( 4 4 4 s s s s s s s R s F +- = +--- =- = λ λ 1 16 1 1 1 16 1 ) 1 ( 4 4 = +- = λ F so λ is recurrent. d. Use ( ) F s λ to calculate E[ T λ ]. Is λ positive recurrent or null recurrent? 2 4 3 4 4 3 4 4 ) 16 1 ( ) 4 1 ( 16 ) 16 1 ( 4 16 1 16 ) ( ' s s s s s s s s s s ds d s F +- +-- +- = +- = λ ∞ < = + = +- +-- +- = = 16 256 / 1 64 / 3 64 / 1 ) 16 1 1 1 ( ) 4 1 1 ( 16 1 ) 16 1 1 1 ( 4 1 ) 1 ( ' ] [ 2 4 3 4 4 3 λ λ F T E so λ is positive recurrent. e. Expand ( ) F s λ in a power series and find f 8 , the probability that “H H T T” first occurs on trial 8. Give a logical explanation for this probability. ( 29 ( 29 1 4 4 4 4 16 1 16 16 1 16 ) (- +- = +- = s s s s s s s F λ Let ( 29 1 4 16 1 ) (- +- = s s s A and expand in a Taylor series by taking derivatives at 0. A (0) = 1 ( 29 ) 4 1 ( 16 1 ) ( ' 3 2 4 s s s s A +- +-- =- so A ’(0) = 1 ( 29 ( 29 ) 4 3 ( 16 1 ) 4 1 ( 16 1 2 ) ( ' ' 2 2 4 2 3 3 4 s s s s s s s A-- +-- +- +- = so A ”(0) = 2 – 0 = 2 ( 29 ( 29 ( 29 ( 29 ) 4 6 ( 16 1 ) 4 3 )( 4 1 ( 16 1 2 ) 4 3 )( 4 1 ( 2 16 1 2 ) 4 1 ( 16 1 6 ) ( 2 4 2 3 3 4 2 3 3 4 3 3 4 4 ) 3 ( s s s s s s s s s s s s s s s A---- +-- +- +- + +- +- + +- +-- = so A (3) (0) = 6 + 0 + 0 – 0 = 6 Simplifying, ( 29 ( 29 ( 29 ) 4 6 ( 16 1 ) 16 4 ( 16 1 18 ) 4 1 ( 16 1 6 ) ( 2 4 5 2 3 4 3 3 4 4 ) 3 ( s s s s s s s s s s s A--- +-- +- +- + +- +-- = ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ) 4 6 ( 16 1 ) 4 6 )( 4 1 ( 16 1 2 ) 16 5 2 ( 16 1 18 ) 16 4 )( 4 1 ( 16 1 54 ) 4 3 ( ) 4 1 ( 3 16 1 6 ) 4 1 ( 16 1 24 ) ( 2 4 3 2 4 4 3 4 5 2 3 3 4 2 2 3 4 4 4 3 5 4 ) 4 (------ +-- +- +- + +- +- + +- +- +-- +- +-- +- +- = s s s s s s s s s s s s s s s s s s s s s s s A So A (4) (0) = 24 – 0 – 0 + 0 + 0 – 6/4 = 45/2...
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This note was uploaded on 10/24/2010 for the course STAT 333 taught by Professor Chisholm during the Spring '08 term at Waterloo.

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STAT_333_Assignment_2_Solutions - STAT 333 Assignment 2...

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