SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

58 CHAPTER 6. DISCRETE-TIME PROCESSES
Case 6.8
S n represents accident history for years n − 1andn
M = {1, 2, 3, 4} as defined in Case 6.8
15. No answer provided.
16. (a)
π 2 =1/4 =0.25
(b) No answer provided.
(c) Under the binomial model with probability of γ = π 2
Pr{Z 15 ≤ 1} =
1∑
j=0
15!
j!(15 − j)! πj 2(1 − π 2 ) 15−j
= 0.9904 when π 2 =0.01
= 0.0802 when π 2 =0.25
For the Markov-chain model we have to look at all possible sample paths:
• Pr{no defectives}
• Pr {1 defective} has several cases:
first item defective
Pr{no defectives} = p 1 (p 11 p 11 ···p 11 )
= π 1 p 14
11 = a
last item defective
Pr = p 2 p 21 p 11 ···p 11 = π 2 p 21 p 13
11 = b
some other item defective
Pr = p 1 p 11 ···p 11 p 12 = π 1 p 13
11p 12 = c
and there are 13 possible “other ones”
Pr = p 1 p 11 p 11 ···p 12 p 21 p 11 p 11 ···p 11
= π 1 p 12
11 p 12p 21 = d