Discrete-time stochastic processes

The non homogeneous poisson process is then given by

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Unformatted text preview: Poisson. Note now that we cannot verify that the two processes are independent from this small increment model. We would have to show that the number of arrivals for process 1 and 2 are independent over (t, t + δ ]. Unfortunately, leaving out the terms of order δ 2 , there is at most one arrival to the original process and no possibility of an arrival to each new process in (t, t + δ ]. If it is impossible for both processes to have an arrival in the same interval, they cannot be independent. It is possible, of course, for each process to have an arrival in the same interval, but this is a term of order δ 2 . Thus, without paying attention to the terms of order δ 2 , it is impossible to demonstrate that the processes are independent. To demonstrate that process 1 and 2 are independent, we ﬁrst calculate the joint PMF for N1 (t), N2 (t) for arbitrary t. Conditioning on a given number of arrivals N (t) for the 2.3. COMBINING AND SPLITTING POISSON PROCESSES 73 original process, we have Pr {N1 (t)=m, N2 (t)=k | N (t)=m+k} = (m + k)! m p (1 − p)k . m!k! (2.22) Equation (2.22) is simply the binomial distribution, since, given m + k arrivals to the original process, each independently goes to process 1 with probability p. Since the event {N1 (t) = m, N2 (t) = k} is a subset of the conditioning event above, Pr {N1 (t)=m, N2 (t)=k | N (t)=m+k} = Pr {N1 (t)=m, N2 (t)=k} . Pr {N (t)=m+k} Combining this with (2.22), we have Pr {N1 (t)=m, N2 (t)=k} = (m + k!) m (∏t)m+k e−∏t p (1 − p)k . m!k! (m + k)! (2.23) (p∏t)m e−∏pt [(1 − p)∏t]k e−∏(1−p)t . m! k! (2.24) Rearranging terms, we get Pr {N1 (t)=m, N2 (t)=k} = This shows that N1 (t) and N2 (t) are independent. To show that the processes are independent, we must show that for any k > 1 and any set of times 0 ≤ t1 ≤ t2 ≤ · · · ≤ tk , the sets {N1 (ti )1 ≤ i ≤ k} and {N2 (tj ); 1 ≤ j ≤ k} are independent of each other. It is equivalent e e to show that the sets {N1 (ti−1 , ti ); 1 ≤ i ≤ k} and {N2 (tj −1 , tj ); 1 ≤ j ≤ k} (where t0 is 0) are independent. The argument above shows this independence for i = j , and for i 6= j , the independence follows from the independent increment property of {N (t); t ≥ 0}. 2.3.2 Examples using independent Poisson processes We have observed that if the arrivals of a Poisson process are split into two new arrival processes, each arrival of the original process independently going into the ﬁrst of the new processes with some ﬁxed probability p, then the new processes are Poisson processes and are independent. The most useful consequence of this is that any two independent Poisson processes can be viewed as being generated from a single process in this way. Thus, if one process has rate ∏1 and the other has rate ∏2 , they can be viewed as coming from a process of rate ∏1 + ∏2 . Each arrival to the combined process then goes to the ﬁrst process with probability p = ∏1 /(∏1 + ∏2 ) and to the second process with probability 1 − p. The above point of view is very useful for ﬁnding probabilities such as Pr {S1k < S2j } where S1k is the epoch of the kth arrival to the ﬁrst process and S2j is the epoch of the j th arrival to the second process. The problem can be rephrased in terms of a combined process to ask: out of the ﬁrst k + j − 1 arrivals to the combined process, what is the probability that k or more of them are switched to the ﬁrst process? (Note that if k or more of the ﬁrst k + j − 1 go to the ﬁrst process, at most j − 1 go to the second, so the kth arrival to the ﬁrst precedes the j th arrival to the second; similarly if fewer than k of the ﬁrst k + j − 1 go to the ﬁrst process, then the j th arrival to the second process precedes the kth arrival 74 CHAPTER 2. POISSON PROCESSES to the ﬁrst). Since each of these ﬁrst k + j − 1 arrivals are switched independently with the same probability p, the answer is Pr {S1k < S2j } = Xk+j −1 i=k (k + j − 1)! pi (1 − p)k+j −1−i . i!(k + j − 1 − i)! (2.25) As an example of this, suppose a queueing system has arrivals according to a Poisson process (process 1) of rate ∏. There is a single server who serves arriving customers in order with a service time distribution F (y ) = 1 − exp[−µy ]. Thus during periods when the server is busy, customers leave the system according to a Poisson process (process 2) of rate µ. Thus, if j or more customers are waiting at a given time, then (2.25) gives the probability that the kth subsequent arrival comes before the j th departure. 2.4 Non-homogeneous Poisson processes The Poisson process, as we deﬁned it, is characterized by a constant arrival rate ∏. It is often useful to consider a more general type of process in which the arrival rate varies as a function of time. A non-homogeneous Poisson process with time varying arrival rate ∏(t) is deﬁned8 as a counting process {N (t); t ≥ 0} which has the independent increment property and, for all t ≥ 0, δ >...
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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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