STAT_333_Assignment_2_Solutions

STAT_333_Assignment_2_Solutions - STAT 333 Assignment 2...

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1. Prove the Delayed Renewal Relation. (Hint: the proof is very similar to that of the Renewal Relation, and you have to be very careful with the starting values of sums.) For n ≥ 1, d n = P(λ occurs on trial n) = ° ( λ occurs first on trial k,and also on trial n) ± =1 = °²λ occurs first on trial k ³ P( λ occurs on trial n|on k) ± =1 = f k P( λ ´ occurs n k trials after λ ) ± =1 = f k r µ n k ± =1 = f k r µ n k ± =0 since f 0 = 0 And λ ²?³ = · ¸ ? ¸ ¸ =0 = 0 + · ¸ ? ¸ ¸ =1 since d 0 = 0 = ( f k r µ n k ± =0 ) ? ¸ ¸ =1 = ( f k r µ n k ± =0 ) ? ¸ ¸ =0 since the n = 0 inner sum is 0 = ( ¹ ¸ ? ¸ )( r µ n s n ± =0 ) ¸ =0 since º· ¸ » = º¹ ¸ » × º?µ ¸ » Thus λ ²?³ = ¼ λ ²?³½ λ ¾ ²?³ . Rearranging gives the required result. 2. A fair 6-sided die is rolled repeatedly. Let λ be the event “the maximum roll so far is ≤ 5” a. Explain carefully why λ is a renewal event. The first waiting time is 1 if the first roll is ≤ 5 and ∞ if it is 6. The between-event waiting time has the exact same distribution. Given it has occurred once, the first roll was ≤ 5. For the maximum to continue to be ≤ 5, the second waiting time is either 1 (if the second roll is ≤ 5) or ∞ if it is 6. All waiting times have the distribution: 1 with prob 5/6 ∞ with prob 1/6 And they are independent. Thus λ is a renewal event. b. Determine the renewal sequence { r n } r 0 = 1 r 1 = P(first roll is ≤ 5) = 5/6 r 2 = P(min is ≤ 5) = P(first and second rolls are ≤ 5) = (5/6) 2 since rolls are independent r n = (5/6) n for n ≥ 1 c. Find f λ by any legitimate method. Is λ recurrent or transient? f
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This note was uploaded on 12/12/2010 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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