Discrete-time stochastic processes

# The problem can be rephrased in terms of a combined

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Unformatted text preview: are independent and both possess the stationary and independent increment properties, it follows from the deﬁnitions that {N (t); t ≥ 0} also possesses the stationary and independent increment properties. Using the approximations in (2.17) for the individual processes, we see that n o n on o e e e Pr N (t, t + δ ) = 0 = Pr N1 (t, t + δ ) = 0 Pr N2 (t, t + δ ) = 0 = (1 − ∏1 δ )(1 − ∏2 δ ) ≈ 1 − ∏δ. n o e where ∏1 ∏2 δ 2 has been dropped. In the same way, Pr N (t, t+δ ) = 1 is approximated by n o e ∏δ and Pr N (t, t + δ ) ≥ 2 is approximated by 0, both with errors proportional to δ 2 . It follows that {N (t), t ≥ 0} is a Poisson process. In the second approach, we have N (t) = N1 (t) + N2 (t). Since N (t), for any given t, is the sum of two independent Poisson rv’s , it is also a Poisson rv with mean ∏t = ∏1 t + ∏2 t. If the reader is not aware that the sum of two independent Poisson rv’s is Poisson, it can be derived by discrete convolution of the two PMF’s (see Exercise 1.18). More elegantly, one can observe that we have already implicitly shown this fact. That is, if we break an interval I into disjoint subintervals, I1 and I2 , the number of arrivals in I (which is Poisson) is the sum of the number of arrivals in I1 and in I2 (which are independent Poisson). Finally, since N (t) is Poisson for each t, and since the stationary and independent increment properties are satisﬁed, {N (t); t ≥ 0} is a Poisson process. In the third approach, X1 , the ﬁrst interarrival interval for the sum process, is the minimum of X11 , the ﬁrst interarrival interval for the ﬁrst process, and X21 , the ﬁrst interarrival interval for the second process. Thus X1 > t if and only if both X11 and X21 exceed t, so Pr {X1 > t} = Pr {X11 > t} Pr {X21 > t} = exp(−∏1 t − ∏2 t) = exp(−∏t). Using the memoryless property, each subsequent interarrival interval can be analyzed in the same way. The ﬁrst approach above was the most intuitive for this problem, but it required constant care about the order of magnitude of the terms being neglected. The second approach was the simplest analytically (after recognizing that sums of independent Poisson rv’s are Poisson), and required no approximations. The third approach was very simple in retrospect, but not very natural for this problem. If we add many independent Poisson processes together, it is clear, by adding them one at a time, that the sum process is again Poisson. What is more interesting is that when many independent counting processes (not necessarily Poisson) are added together, the sum process often tends to be approximately Poisson if the individual processes have small rates compared to the sum. To obtain some crude intuition about why this might be expected, note that the interarrival intervals for each process (assuming no bulk arrivals) will tend to be large relative to the mean interarrival interval 72 CHAPTER 2. POISSON PROCESSES for the sum process. Thus arrivals that are close together in time will typically come from diﬀerent processes. The number of arrivals in an interval large relative to the combined mean interarrival interval, but small relative to the individual interarrival intervals, will be the sum of the number of arrivals from the diﬀerent processes; each of these is 0 with large probability and 1 with small probability, so the sum will be approximately Poisson. 2.3.1 Subdividing a Poisson process Next we look at how to break {N (t), t ≥ 0}, a Poisson counting process of rate ∏, into two processes, {N1 (t), t ≥ 0} and {N2 (t), t ≥ 0}. Suppose that each arrival in {N (t), t ≥ 0} is sent to the ﬁrst process with probability p and to the second process with probability 1 − p (see Figure 2.6). Each arrival is switched independently of each other arrival and independently of the arrival epochs. We shall show that the resulting processes are each Poisson, with rates ∏1 = ∏p and ∏2 = ∏(1 − p) respectively, and that furthermore the two processes are independent. Note that, conditional on the original process, the two new processes are not independent; in fact one completely determines the other. Thus this independence might be a little surprising. N (t) rate ∏ ✟ ✟✟ p ✲✟✟ ❍❍ ❍1−p ❍❍ N1 (t) rate ∏1 = p∏ ✲ N2 (t) ✲ rate ∏2 = (1−p)∏ Figure 2.6: Each arrival is independently sent to process 1 with probability p and to process 2 otherwise. First consider a small increment (t, t + δ ]. The original process has an arrival in this incremental interval with probability ∏δ (ignoring δ 2 terms as usual), and thus process 1 has an arrival with probability ∏δ p and process 2 with probability ∏δ (1 − p). Because of the independent increment property of the original process and the independence of the division of each arrival between the two processes, the new processes each have the independent increment property, and from above have the stationary increment property. Thus each process is...
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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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