Discrete-time stochastic processes

11 density for the order statistics of an iid 2

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Unformatted text preview: function G(y ); the service time is the interval from start to completion of service and is also independent of arrival epochs. We would like to find the distribution function of the number of customers being served at a given epoch τ . Let {N (t); t ≥ 0} be the Poisson counting process of customer arrivals. Consider the arrival times of those customers that are still in service at some fixed time τ . In some arbitrarily small interval (t, t + δ ], the probability of an arrival is δ ∏ + o(δ ) and the probability of 2 or more arrivals is negligible (i.e., o(δ )). The probability that an arrival occurred in (t, t + δ ] and that that customer is still being served at time τ > t is then δ ∏[1 − G(τ − t)] + o(δ ). Consider a counting process {N1 (t); 0≤t≤τ } where N1 (t) is the number of arrivals between 0 and t that are still in service at τ . This counting process has the independent increment property. To see this, note that the overall arrivals in {N (t); t ≥ 0} have the independent increment property; also the arrivals in {N (t); t ≥ 0} have independent service times, and thus are independently in or not in {N1 (t); 0 ≤ t < τ }. It follows that {N1 (t); 0 ≤ t < τ } is a non-homogeneous Poisson process with rate ∏[1 − G(τ − t)] at time t ≤ τ . The expected number of arrivals still in service at time τ is then Zτ Zτ m(τ ) = ∏ [1 − G(τ − t)] dt = ∏ [1 − G(t)] dt. (2.32) t=0 t=0 and the PMF of the number in service at time τ is given by Pr {N1 (τ ) = n} = m(τ )n exp(−m(τ )) . n! (2.33) Note that as τ → 1, the integral in (2.32) approaches the mean of the service time distribution (i.e., it is the integral of the complementary distribution function, 1 − G(t), of the 2.5. CONDITIONAL ARRIVAL DENSITIES AND ORDER STATISTICS 77 ✲ N1 (τ ) = Customers in service at τ ✟✟G(τ −t) ✟1 − N (t) ✲✟✟ ❍❍ ❍G(τ −t) ❍❍ ✲ N (τ ) − N1 (τ ) = Customers departed by τ Figure 2.8: Poisson arrivals {N (t); t ≥ 0} can be considered to be split in a nonhomogeneous way. An arrival at t is split with probability 1 − G(τ − t) into a process of customers still in service at τ . service time). This means that in steady-state (as τ → 1), the distribution of the number in service at τ depends on the service time distribution only through its mean. This example can be used to model situations such as the number of phone calls taking place at a given epoch. This requires arrivals of new calls to be modeled as a Poisson process and the holding time of each call to be modeled as a random variable independent of other holding times and of call arrival times. Finally, as shown in Figure 2.8, we can regard {N1 (t); 0≤t ≤ τ } as a splitting of the arrival process {N (t); t≥0}. By the same type of argument as in Section 2.3, the number of customers who have completed service by time τ is independent of the number still in service. 2.5 Conditional arrival densities and order statistics A diverse range of problems involving Poisson processes are best tackled by conditioning on a given number n of arrivals in the interval (0, t], i.e., on the event N (t) = n. Because of the incremental view of the Poisson process as independent and stationary arrivals in each incremental interval of the time axis, we would guess that the arrivals should have some sort of uniform distribution given N (t) = n. More precisely, the following theorem shows that the joint density of S (n) = (S1 , S2 , . . . , Sn ) given N (t) = n is uniform over the region 0 < S1 < S2 < · · · < Sn < t. Theorem 2.6. Let fS(n) |N (t) (s(n) | n) be the joint density of S(n) conditional on N (t) = n. This density is constant over the region 0 < s1 < · · · < sn < t and has the value fS(n) |N (t) (s(n) | n) = n! . tn (2.34) Two proofs are given, each illustrative of useful techniques. Proof 1: Recall that the joint density of the first n + 1 arrivals S n+1 = (S1 . . . , Sn , Sn+1 with no conditioning is given in (2.14). We first use Bayes law to calculate the joint density of S n+1 conditional on N (t) = n. fS (n+1) |N (t) (s (n+1) | n) pN (t) (n) = pN (t)|S (n+1) (n|s (n+1) )fS (n+1) (s (n+1) ). Note that N (t) = n if and only if Sn ≤ t and Sn+1 > t. Thus pN (t)|S (n+1) (n|s (n+1) ) is 1 if Sn ≤ t and Sn+1 > t and is 0 otherwise. Restricting attention to the case N (t) = n, Sn ≤ t 78 CHAPTER 2. POISSON PROCESSES and Sn+1 > t, fS (n+1) |N (t) (s (n+1) | n) = = = fS (n+1) (s (n+1) ) pN (t) (n) ∏n+1 exp(−∏sn+1 ) (∏t)n exp(−∏t) /n! n!∏ exp(−∏(sn+1 − t) . tn (2.35) This is a useful expression, but we are interested in S (n) rather than S (n+1) . Thus we break up the left side of (2.35) as follows: fS (n+1) |N (t) (s (n+1) | n) = fS (n) |N (t) (s (n) | n) fSn+1 |S (n) N (t) (sn+1 |s (n) , n). Conditional on N (t) = n, Sn+1 is the first arrival epoch after t, which by the memoryless property is independent of S n . Thus that final term is simply ∏ exp(−∏(sn+1 − t)) for sn+1 > t. Substituting this into (2.35), the r...
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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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