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Chapt 10c

# Chapt 10c - 10.5 Poisson Process 353 restaurant will go to...

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10.5 Poisson Process 353 restaurant will go to refuel at the gas station before going back to the highway with a probability of 0.25. What is the probability that exactly 10 cars have been refueled at the gas station within a particular two-hour period? Solution The process that governs car arrivals at the gas station is Poisson with a rate of λ G = p λ = ( 0 . 25 )( 12 ) = 3 cars per hour. Thus, if K represents the number of cars that arrive at the gas station within 2 hours, the probability that K = 10 cars is given by P [ K = 10 ] = ( 2 λ G ) 10 10 ! e 2 λ G = 6 10 10 ! e 6 = 0 . 0413 trianglesolid 10.5.11 Random Incidence Consider a Poisson process X ( t ) in which events (or arrivals) occur at times T 0 = 0 , T 1 , T 2 ,... . Let the interarrival times Y k be defined as follows: Y 1 = T 1 T 0 Y 2 = T 2 T 1 ... Y k = T k T k 1 These interarrival times are illustrated in Figure 10.6, where A k denotes the k th arrival. The Poisson process belongs to a class of random processes called renewal processes that have the property that the Y k are mutually independent and iden- tically distributed. For the Poisson process with mean arrival rate λ , the Y k are exponentially distributed with mean 1 , as discussed earlier. Consider the following problem in connection with the Y k . Assume the T k are the points in time that buses arrive at a bus stop. A passenger arrives at the bus stop at a random time and wants to know how long he or she will wait until the next bus arrival. This problem is usually referred to as the random incidence problem , since the subject (or passenger in this example) is incident to the process Figure 10.6 Interarrival Times of a Poisson Process

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354 Chapter 10 Some Models of Random Processes Figure 10.7 Random Incidence at a random time. Let R be the random variable that denotes the time from the moment the passenger arrived until the next bus arrival. R is referred to as the residual life of the Poisson process. Also, let W denote the length of the interar- rival gap that the passenger entered by random incidence. Figure 10.7 illustrates the random incidence problem. Let f Y ( y ) denote the PDF of the interarrival times; let f W ( w ) denote the PDF of W , the gap entered by random incidence; and let f R ( r ) denote the PDF of the residual life, R . The probability that the random arrival occurs in a gap of length between w and w + dw can be assumed to be directly proportional to the length w of the gap and relative occurrence f Y ( w ) dw of such gaps. That is, f W ( w ) dw = β wf Y ( w ) dw where β is the constant of proportionality. Thus, f W ( w ) = β wf Y ( w ) . Since f W ( w ) is a PDF, we have that integraldisplay −∞ f W ( w ) dw = 1 = β integraldisplay −∞ wf Y ( w ) dw = β E [ Y ] Thus, β = 1 / E [ Y ] , and we obtain f W ( w ) = wf Y ( w ) E [ Y ] The expected value of W is given by E [ W ] = E bracketleftbig Y 2 bracketrightbig / E [ Y ] . This result applies to all renewal processes. For a Poisson process, Y is exponentially distributed with E [ Y ] = 1 and E bracketleftbig Y 2 bracketrightbig = 2 2 . Thus, for a Poisson process we obtain f W ( w ) = λ wf Y ( w ) = λ 2 we λ w w 0 E [ W ] = 2
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