Discrete-time stochastic processes

Discrete-time stochastic processes

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Unformatted text preview: j such that Pkj > 0. It is not hard to see that if the distribution of inter-renewal intervals for one value of j is arithmetic with span d, then the distribution of inter-renewal intervals for each i is arithmetic with the same span (see exercise 5.19). 5.8. EXAMPLE — THE M/G/1 QUEUE 227 For the example of Figure 5.8, we see by inspection that U (1) = 5.5 and U (2) = 1. Thus p1 = 11/13, and p2 = 2/13. For a semi-Markov process, knowing the probability that X (t) = j for large t does not completely specify the steady-state behavior. Another important steady state question is to determine the fraction of time involved in i to j transitions. To make this notion precise, define Y (t) as the residual time until the next transition after time t (i e., t + Y (t) is the epoch of the next transition after time t). We want to determine the fraction of time t over which X (t) = i and X (t + Y (t)) = j . Equivalently, for a non-arithmetic process, we want to determine Pr {X (t) = i, X (t + Y (t) = j )} in the limit as t → 1. Call this limit Q(i, j ). Consider a renewal process, starting in state i and with renewals on transitions to state i. Define a reward R(t) = 1 for X (t) = i, X (t + Y (t)) = j and R(t) = 0 otherwise (see Figure 5.10). That is, for each n such that X (Sn ) = i and X (Sn+1 ) = j , R(t) = 1 for Sn ≤ t < Sn+1 . The expected reward in an inter-renewal interval is then Pij U (i, j ) . It follows that Q(i, j ) is given by Rt R(τ )(dτ Pij U (i, j ) pi Pij U (i, j ) Q(i, j ) = lim 0 = = . (5.69) t→1 t W (i) U (i) t t ✛ Xn =i ✛ Un ✲ t Xn+1 =j t Xn+2 6=j Wk t t Xn+3 =i Xn+4 6=j ✲ t Xn+5 =i t Xn+6 =j Figure 5.10: The renewal-reward process for i to j transitions. The expected value of Un if Xn = i and Xn+1 = j is U (i, j ) and the expected interval between entries to i is W (i). 5.8 Example — the M/G/1 queue As one example of a semi-Markov chain, consider an M/G/1 queue. Rather than the usual interpretation in which the state of the system is the number of customers in the system, we view the state of the system as changing only at departure times; the new state at a departure time is the number of customers left behind by the departure. This state then remains fixed until the next departure. New customers still enter the system according to the Poisson arrival process, but these new customers are not considered as part of the state until the next departure time. The number of customers in the system at arrival epochs does not in general constitute a “state” for the system, since the age of the current service is also necessary as part of the statistical characterization of the process. One purpose of this example is to illustrate that it is often more convenient to visualize the transition interval Un = Sn − Sn−1 as being chosen first and the new state Xn as being 228 CHAPTER 5. COUNTABLE-STATE MARKOV CHAINS chosen second rather than choosing the state first and the transition time second. For the M/G/1 queue, first suppose that the state is some i > 0. In this case, service begins on the next customer immediately after the old customer departs. Thus, Un , conditional on Xn = i for i > 0, has the distribution of the service time, say G(u). The mean interval until a state transition occurs is Z1 U (i) = [1 − G(u)]du; i > 0. (5.70) 0 Given the interval u for a transition from state i > 0, the number of arrivals in that period is a Poisson random variable with mean ∏u, where ∏ is the Poisson arrival rate. Since the next state j is the old state i, plus the number of new arrivals, minus the single departure, Pr {Xn+1 = j | Xn = i, Un = u} = (∏u)j +i+1 exp(−∏u) . (j − i + 1)! (5.71) for j ≥ i − 1. For j < i − 1, the probability above is 0. The unconditional probability Pij of a transition from i to j can then be found by multiplying the right side of (5.71) by the probability density g (u) of the service time and integrating over u. Z1 G(u)(∏u)j −i+1 exp(−∏u) Pij = du; j ≥ i − 1, i > 0. (5.72) (j − i + 1) 0 For the case i = 0, the server must wait until the next arrival before starting service. Thus the expected time from entering the empty state until a service completion is Z1 U (0) = (1/∏) + [1 − G(u)]du. (5.73) 0 We can evaluate P0j by observing that the departure of that first arrival leaves j customers in this system iff j customers arrive during the service time of that first customer; i.e., the new state doesn’t depend on how long the server waits for a new customer to serve, but only on the arrivals while that customer is being served. Letting g (u) be the density of the service time, Z1 g (u)∏u)j exp(−∏u) P0j = du; j ≥ 0. (5.74) j! 0 5.9 Summary This chapter extended the finite-state Markov chain results of Chapter 4 to the case of countably-infinite state spaces. It also provided an excellent example of how renewal processes can be used for understanding other kinds of processes. In Section 5.1, the firstpassage-time random variables were used to construct renewal processes with renewals on successive transitio...
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This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.

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