STAT_333_Assignment_2Solutions

# STAT_333_Assignment_2Solutions - STAT 333 Assignment 2...

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STAT 333 Assignment 2 SOLUTIONS Due: Friday, March 5 at the beginning of class 1. Suppose we toss a fair coin repeatedly. 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 0 = 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 () Rs (the generating function of { r n }), use it to obtain Fs and prove recurrence. , 1 16 1 ) ( 1 ) ( 4 4 4 2 1 0 s s s s r s R n n n n n 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 1 16 1 1 1 16 1 ) 1 ( 4 4 F so λ is recurrent. d. Use 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 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 F T E so λ is positive recurrent. e. Expand 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.     1 4 4 4 4 16 1 16 16 1 16 ) ( s s s s s s s F Let   1 4 16 1 ) ( s s s A and expand in a Taylor series by taking derivatives at 0. A (0) = 1   ) 4 1 ( 1 ) ( ' 3 2 4 s s s s A so A ’(0) = 1     ) 4 3 ( 1 ) 4 1 ( 1 2 ) ( ' ' 2 2 4 2 3 3 4 s s s s s s s A so A ”(0) = 2 – 0 = 2

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        ) 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,       ) 4 6 ( 16 1 ) 4 ( 1 18 ) 4 1 ( 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             ) 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 (Mathematical software use for these derivatives is allowed. Can also be done with MATH239- style series expansion techniques instead of derivatives) Now we have A ( s ) = 1 + (1/1!) s + (2/2!) s 2 + (6/3!) s 3 + (45/2/4!) s 4 + … And thus F λ ( s ) = ( s 4 /16) A ( s ) = (1/16) s 4 + (1/16) s 5 +(1/16) s 6 +(1/16) s 7 + (15/256) s 8 + … So f 8 = 15/256.
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STAT_333_Assignment_2Solutions - STAT 333 Assignment 2...

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