renewal1 practice problems

# renewal1 practice problems - Stat 333 Summer 2007 Renewal...

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Stat 333 Summer 2007 Renewal theory: Practice Problems 1. Below are some hypothetical frst waiting-time distributions f n = P ( T λ = n )whe re λ is a renewal event oF a stochastic process. In each case determine whether λ is transient or recurrent; iF recurrent, Further determine whether λ is positive recurrent or null recurrent by computing E ( T λ ). (a) P ( T λ = n )=(2 / 3) n , n =3 , 4 , 5 ... (b) P ( T λ = n )=(1 / 2) n , n =1 , 2 , 3 , ... (c) Let c = n =4 1 /n 3 / 2 . Using any legitimate arguments you might have From calculus (and you do have them), determine whether c is fnite or infnite. IF c is fnite and you defne P ( T λ = n )=1 / ( cn 3 / 2 )For n =4 , 5 , 6 , ... , determine whether λ is transient, positive recurrent, or null recurrent. Solution: (a) P ( T λ = n )=( 2 3 ) n for k , 4 ,... P ( T λ < )= ± n =3 ( 2 3 ) n = (2 / 3) 3 1 2 / 3 = 8 9 < 1 , so λ is transient. (b) P ( T λ = n )=2 n for n , 2 , 3 P ( T λ < ± n =1 P ( T λ = n ± n =1 2 n = 1 / 2 1 / 2 . So f λ = P ( T λ < )=1,then λ is recurrent. E ( T λ ± n =1 n 2 n = E (geometric distribution with p / 2) = 1 1 / 2 =2 . So λ is positive recurrent. (c) ±irstly, we introduce a Famous result For series: ± n =4 1 n p = ² iF p 1 ±inite number < iF p> 1 . ± romtheaboveresu lt ,weknow c is a fnite number. Since P ( T λ < ³ ± n =4 1 n 3 / 2 ´µ c , then λ is recurrent. Note that E ( T λ c 1 ± n =4 n · 1 n (3 / 2) = c 1 ± n =4 1 n (1 / 2) = , then λ is null recurrent. 1

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2. Suppose λ is a renewal event and V λ = total number of occurrences of λ in the sequence. (a) If E ( V λ ) = 7, Fnd the probability that λ occurs at least once in the sequence. Is λ recurrent or transient? (b) If E ( V λ ) = 7, what is the probability of exactly 5 occurrences of λ in the sequence? Of at least 5 occurrences? (Hint: use the probability distribution of V λ .) (c) ±or a general transient event λ , determine the smallest value of f λ which will yield E ( V λ ) 50. Solution: (a) E ( V λ )= f λ 1 f λ =7 ( finite, so transient ) f λ =7 7 f λ , so 8 f λ . So f λ / 8 < 1, then λ is transient. (b) P ( V λ =5)= f 5 λ (1 f λ )=(7 / 8) 5 (1 / 8) = 0 . 0641 P ( V λ 5) = f 5 λ =(7 / 8) 5 = 0 . 5129 . (c) E ( V λ f λ 1 f λ 50 51 f λ 50 f λ 50 51 .
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renewal1 practice problems - Stat 333 Summer 2007 Renewal...

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