Problem Set 8

Transcription

1 Eli H. Ross Alberto De Sole November, 8.5 Problem Set 8 Exercise 36 Let X t and Y t be two independent Poisson processes with rate parameters λ and µ respectively, measuring the number of customers arriving in shops and, respectively. a) What is the probability that a customer arrives in store before any customer arrives in store? b) Show that Z t = X t + Y t is a Poisson process, and compute its rate. c) What is the probability that in the first four hours a total of customers have arrived in the two stores? d) Given that exactly customers have arrived at the two stores, what is the probability that the all went to store? e) Let T ) be the time of arrival of the first customer in store. Then X ) is the T number of customers in store at the time the first customer arrives in store. Find the probability distribution of the random variable X ). T f) Let T be the first time that at least one customer has arrive in each of the two shops. Find the probability density function for T. a) The probability that the first customer arrives in store at time x is λe xλ. The probability that the first customer arrives in store at time y is µe yµ. Hence, the probability that the first customer arrives in store at time x and the first customer arrives in store at time y is λµe xλ yµ. The desired probability is the following integral: which we can evaluate as x λµe xλ yµ dy dx, λe xλ+µ) dx = λ λ + µ. b) Let τ Z > k be the event that a waiting time for Z t greater than k. Note that P[τ Z > k] = P[τ X > k, τ Y > k]. Since X t and Y t are independent, this is P[τ X > k] P[τ Y > k] = e λk e µk = e λ+µ)k. Hence, Z t is a Poisson process with rate λ + µ.

2 c) The probability that n customers have arrived at store during the hours is λ)n e λ and the probability that n customers have arrived at store during the hours is µ)n e µ n!. We sum over all of the possibilities: λ) n e λ n= n! =! e λ+µ) µ) n e µ ) = e λ+µ) n)! n= n= λ n µ n n! n)! ) )λ n µ n = 3 n 3 e λ+µ) λ + µ). n! d) The desired conditional probability is λ) e λ! µ) e µ )! 3 3 e λ+µ) λ + µ) = λ λ + µ) = ) λ. λ + µ e) See that = P[X T ) = k] = P[X T ) ) P[X t = k] P[T ) = t] dt = ) = k T ) = t] P[T ) = t] dt ) λt) k e λt µe tµ dt k! = λk µ λ + µ) k+. Thus, the probability distribution for P[X ) T and elsewhere. = k] is ) λ k µ λ + µ λ + µ for k =,,,..., f) The probability that the first customer arrives in store at time x is λe xλ. The probability that the first customer arrives in store at time y is µe yµ. Hence, the probability that the first customer arrives in store at time x and the first customer arrives in store at time y is λµe xλ yµ. If T = t, then either the second store to get a customer was store at time t or store at time t. Thus, we have = P[T = t] = t λµe tλ yµ dy + t λµe xλ tµ dx λe tλ+µ) + λe tλ) + µe tλ+µ) + µe tµ) = λe tλ + µe tµ λ + µ)e λ+µ)t.

3 Exercise 37 Consider the continuous time Markov chain with state space S = {,, 3, } and infinitesimal generator A = a) Find the equilibrium distribution π. 3 3 b) Suppose the chain starts in state. What is the expected amount of time until it changes for the first time? c) Again assume that the chain starts in state. What is the expected amount of time until the chain reaches state?. a) Take πa = where π = [π π π 3 π ]. Expanding gives 3π + π 3 = π 3π + π 3 = π + π π 3 + π = π + π + π 3 π =. We have π 3 = 3π, so π 3π + 6π =, which gives π = 7 3 π. Finally, π π + 3π π =, so π = 9 3 π. Hence, i= π i = 38 3 π =, so π = Back substitution gives the complete equilibrium distribution: π = [ ]. b) The change rate is 3 per unit of time. Hence, the expected amount of time until the first jump is 3 of a unit of time. c) The expected time of first passage for state is τ ). Note that τ ) = A )), where A ) is the matrix obtained from A by erasing row and column. Thus, we have that 3 A ) = 3 so A )) = 8/9 6/9 5/9 /9 /9 6/9 3/9 7/9 9/9 Hence, A )) ) = [ ], so τ =. 3.

4 Exercise 38 Consider the birth and death process with λ n = /n + ) and µ n =. Show that the process is recurrent and admits and equilibrium distribution. Find the equilibrium distribution. Since there is no absorbing state, the process will be recurrent. Indeed, see that from each n = k, we can get to n = k with positive parameter, and we can get to n = k + with positive parameter /n + ). At, we will either grow or not grow. Clearly, then, we can get from any state to any other state, so the process is recurrent. There must be an equilibrium distribution because λ n...λ n= µ n...µ <. To see this, note that the denominator is, and the numerator is n! ; hence, the sum is e. Let A be the standard birthdeath process infinitesimal generator, with the specific values as given in this problem. Let π = [π π π...]. Then, we want πa =. Expanding this out gives: for the first term, and then λ π + µ π = λ n π n λ n + µ n ) π n + µ n+ π n+ = for all the other terms n =,,...). We substitute µ n = and λ n = /n + ) to obtain and π + π = ) n π n n + + π n + π n+ = for n =,,..., so π n+ = n+ n+π n n π n. Inserting π = π gives π = π. Then, π 3 = 3 π π = 6 π. Then, π = 5 6 π 3 π = π. We have compelling evidence that π n = n! π. Let s prove this using induction. We ve shown the base case up to n =, in fact); now, suppose that this holds for n =,,,..., m. Then, π m+ = m + m + m! π m m + = π m + )! ) m! m + m )! π = π m! m + ) = π m + ) m + ) m + )! m )! m ) = π m + )!, which completes the induction. Now, we need n= π n = π n= n! =, so π e =, so π = e. Hence, the equilibrium distribution is π = [π π π...] where π n = e n!. )

5 Exercise 39 K&T. p.3) A new product a Home Helicopter to solve the commuting problem) is being introduced. The sales are expected to be determined by both media advertising and word-of-mouth advertising. Assume that media advertising creates new customers according to a Poisson process of rate α = customer per month. For the word-of-mouth advertising, assume that each purchaser of a Home Helicopter will generate new customers at a rate θ = customers per month. Let X t be the total number of Home Helicopter customers up to time t. a) Model X t as a pure birth process by specifying the birth parameters λ n, n =. b) What is the probability that exactly two helicopters are sold during the first month? a) There is birth rate from media advertising regardless of the number of customers), and birth rate from each customer. Hence, we can model this as a pure birth process with parameter λ n = n + for n =,,,... b) Using the equations given in the textbook, we have that and for n =,,... For P t), we have λ =, so P t) = λ P t) P nt) = λ n P n t) + λ n P n t) P t) = P t) which gives P t) = Ce t. The initial condition of P ) = gives C =, so P t) = e t. Then, since λ = + = 3, we have P t) = 3P t) + e t. Multiplying by the integrating factor e 3t gives e 3t P t) = e t dt = et + C, so P t) = e t + Ce 3t. Using the condition that P ) =, we find C =, so P t) = e t e 3t. Finally, we have λ = + = 5, so P t) = 5P t) + 3 e t e 3t). Multiplying by the integrating factor e 5t gives 3 e 5t P t) = et 3 ) et dt = 3 8 et 3 et + C, so P t) = 3 8 e t 3e 3t + Ce 5t. Using the condition that P ) =, we find C = 8 3, so P t) = 8 3e t + 3 e 3t + 8 3e 5t. Our desired value is P ) = 3 8e 3 e e 5. 5

6 Exercise K&T. p.3) Consider a pure birth process on the states,,..., N, with birth parameters λ n = N n)λ for n =,,..., N. Suppose that X =. Determine P n t) = P[X t = n] for n =,,. Using the equations given in the textbook, we have that and P t) = λ P t) P nt) = λ n P n t) + λ n P n t) for n =,,... For P t), we have λ = Nλ, so P t) = NλP t) P t) P t) = Nλ ln P t)) = Nλt + C so P t) = Ke Nλt. The initial condition of P ) = gives K =, so P t) = e Nλt. Then, P t) = λ P t) + λ P t) = N )λp t) + Nλe Nλt. We multiply by the integrating factor e N )λt to obtain e N )λt P t) = Nλe λt dt = Ne λt + C, so P t) = Ne Nλt + Ce N )λt. Using the condition P ) =, we find that C = N. Thus, P t) = N e N )λt e Nλt) ), or P t) = Ne Nλt e λt. Finally, we have that P t) = λ P t) + λ P t) = N )λp t) + N )λ Ne Nλt e λt ). We multiply by the integrating factor e N )λt to obtain e N )λt P t) = = NN )e λt + N )λn e λt e λt)) dt NN ) e λt + C, so P t) = NN )e N )λt + NN ) e Nλt + Ce N )λt. Using the condition P ) =, we find that C = NN ). Thus, P t) = NN )e N )λt + NN ) e Nλt + NN ) e N )λt. We can factor this nicely as P t) = NN ) ) e Nλt e λt e λt +. 6