Discrete-time stochastic processes

n tk1 t renaming t as tk and t0 as tk1 we see

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Unformatted text preview: h(x) + h(t) for all x, t ≥ 0. These two statements (see Exercise 2.6) imply that h(x) must be linear in x, and Pr {X > x} must be exponential in x. Although the exponential distribution is the only memoryless distribution, it is interesting to note that if we restrict the definition of memoryless to integer times, then the geometric distribution is memoryless, so the Bernoulli process in this respect seems like a discrete-time version of the Poisson process. We now use the memoryless property of the exponential rv to find the distribution of the first arrival in a Poisson process after some given time t > 0. We not only find this distribution, but also show that this first arrival after t is independent of all arrivals up to and including t. Note that t is an arbitrarily selected constant here; it is not a random variable. Let Z be the duration of the interval from t until the first arrival after t. First we find Pr {Z > z | N (t) = 0} . 62 CHAPTER 2. POISSON PROCESSES ✛ ✛ ✛ X2 ✲ ✲ ✲ Z X1 0 t S1 S2 Figure 2.2: For some fixed t, consider the event N (t) = 0. Conditional on this event, Z is the interval from t to S1 ; i.e., Z = X1 − t. As illustrated in Figure 2.2, for {N (t) = 0}, the first arrival after t is the first arrival of the process. Stating this more precisely, the following events are identical:3 {Z > z } \ \ {N (t) = 0} = {X1 > z + t} {N (t) = 0}. The conditional probabilities are then Pr {Z > z | N (t)=0} = Pr {X1 > z + t | N (t)=0} = Pr {X1 > z + t | X1 > t} −∏z = Pr {X1 > z } = e . (2.5) (2.6) In (2.5), we used the fact that {N (t) = 0} = {X1 > t}, which is clear from Figure 2.1. In (2.6) we used the memoryless condition in (2.4) and the fact that X1 is exponential. Next consider the condition that there are n arrivals in (0, t] and the nth occurs at epoch Sn = τ ≤ t. The argument here is basically the same as that with N (t) = 0, with a few extra details (see Figure 2.3). ✛ X3 ✲ N (t) ✛ X2 ✲ ✻ ✛ Z ✲ ✛ X1 ✲ 0 S1 τ S2 t S3 Figure 2.3: Given N (t) = 2, and S2 = τ , X3 is equal to Z + (t − τ ). Also, the event {N (t)=2, S2 =τ } is the same as the event {S2 =τ , X3 >t−τ }. Conditional on N (t) = n and Sn = τ , the first arrival after t is the first arrival after the arrival at Sn , i.e., Z = z corresponds to Xn+1 = z + t − τ . Stating this precisly, the following events are identical: \ \ \ \ {Z > z } {N (t) = n} {Sn = τ } = {Xn+1 > z +t−τ } {N (t) = n} {Sn = τ }. 3 It helps intuition to sometimes think of one event A as conditional on another event B . More precisely, T A given B is the set of sample points in B that are also in A, which is simply A B . 2.2. DEFINITION AND PROPERTIES OF THE POISSON PROCESS 63 Note that Sn = τ is an event of zero probability, but Sn is a sum of n IID random variables with densities, and thus has a density itself, so that other events can be conditioned on it. Pr {Z > z | N (t)=n, Sn =τ } = Pr {Xn+1 > z +t−τ | N (t)=n, Sn =τ } = Pr {Xn+1 > z +t−τ | Xn+1 >t−τ , Sn =τ } = Pr {Xn+1 > z +t−τ | Xn+1 >t−τ } −∏z = Pr {Xn+1 > z } = e . (2.7) (2.8) (2.9) (2.10) In (2.8), we have used the fact that, given Sn = τ , the event N (t) = n is the same as Xn+1 > t − τ (see Figure 2.3). In (2.9) we used the fact that Xn+1 is independent of Sn . In (2.10) we used the memoryless condition in (2.4) and the fact that Xn+1 is exponential. The same argument applies if, in (2.7), we condition not only on Sn but also on S1 , . . . , Sn−1 . Since this is equivalent to conditioning on N (τ ) for all τ in (0, t], we have Pr {Z > z | {N (τ ), 0 < τ ≤ t}} = exp(−∏z ). (2.11) The following theorem states this in words. Theorem 2.1. For a Poisson process of rate ∏, and any given time t > 0, the interval from t until the first arrival after t is a nonnegative rv Z with the distribution function 1 − exp[−∏z ] for z ≥ 0. This rv is independent of al l arrival epochs before time t and independent of N (τ ) for al l τ ≤ t. The length of our derivation of (2.11) somewhat hides its conceptual simplicity. Z , conditional on the time τ of the last arrival before t, is simply the remaining time until the next arrival, which, by the memoryless property, is independent of τ ≤ t, and hence also independent of everything before t. Next consider subsequent interarrival intervals after a given time t. For m ≥ 2, let Zm be the interarrival interval from the m − 1st arrival epoch after t to the mth arrival epoch after t. Given N (t) = n, we see that Zm = Xm+n , and therefore Z2 , Z3 , . . . , are IID exponentially distributed random variables, conditional on N (t) = n (see Exercise 2.8). Let Z in (2.11) become Z1 here. Since Z1 is independent of Z2 , Z3 , . . . and independent of N (t), we see that Z1 , Z2 , . . . are unconditionally IID and also independent of N (t). It should also be clear that Z1 , Z2 , . . . are independent of {N (τ ); 0 < τ ≤ t}. The...
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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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