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# 416-6 - = 1 2 for the remaining two-thirds of its clients...

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Denker Spring 2010 416 Stochastic Modeling - Assignment 6 Due Date: Wednesday, March 3, 2010 (exception) Problem 1: (Problem 44, p. 271) Suppose that a population consists of a Fxed number of genes, say m genes. If any generation has exactly i of its m genes of type 1, then the next generation will have j type 1 genes (and hence m - j type 2 genes) with probability p m j P i j ( m - i ) m - j m - m . Let X n denote the number of type 1 genes in the n-th generation, and assume that X 0 = i . 1. ±ind E [ X n ]. 2. What is the probability that eventually all the genes will be type 1? Problem 2: (Problem 47, p.272) Let ( X n ) n 0 denote an ergodic irreducible Markov chain with limiting probabilities π j . DeFne the process ( Y n ) n 1 by Y n = ( X n - 1 , X n ) . Explain why this new process is a Markov chain and Fnd its limiting probabilities (if they exist). Problem 3: (Problem 53, p. 273) ±ind the average premium received per policyholder of the insurance company of Example 4.23 if λ = 1 / 4 for one-third of its clients, and λ
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Unformatted text preview: = 1 / 2 for the remaining two-thirds of its clients. Problem 4: (Problem 56, p.271) Suppose a gambler wins in a sequence of i.i.d. trials with success probability (i.e. to win) p . The gambler continues betting until he wins n or loses m trials. What is the probability that the gambler quits a winner? Problem 5: (Problem 52, p. 272) A taxi driver provides service to two zones A and B in a city. ±ares picked up in zone A will have destination in zone A with probability 0.6 or in zone B with probability 0.4. ±ares picked up in zone B will have destination in A with probability 0.3 or in zone B with probability 0.7. The driver’s expected proFt for a trip entirely in zone A is \$6, for a trip entirely in zone B it is \$8; and for a trip between the two zones it is \$12. ±ind the drivers average proFt per trip. What is the steady state of this Markov chain?...
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