THE POISSON PROCESS

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3 t 0 (111)

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P {Sk t} = 1

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The Erlang (k, ) distribution has the probability density k t k 1 e t /(k 1)! Theorem 111 For any t > 0, P {N (t) = k} = e t ( t)k , k! k = 0, 1, (112)

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That is, N (t) is Poisson distributed with mean t Proof The proof is based on the simple but useful observation that the number of arrivals up to time t is k or more if and only if the kth arrival occurs before or at time t Hence P {N (t) k} = P {Sk t}

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The result next follows from P {N (t) = k} = P {N (t) k} P {N (t) k + 1} The following remark is made To memorize the expression (111) for the distribution function of the Erlang (k, ) distribution it is easiest to reason in reverse order: since the number of arrivals in (0, t) is Poisson distributed with mean t and the kth arrival time Sk is at or before t only if k or more arrivals occur in (0, t), it follows that P {Sk t} = j =k e t ( t)j /j ! The memoryless property of the Poisson process Next we discuss the memoryless property that is characteristic for the Poisson process For any t 0, de ne the random variable t as t = the waiting time from epoch t until the next arrival The following theorem is of utmost importance Theorem 112 For any t 0, the random variable t has the same exponential distribution with mean 1/ That is, P { t x} = 1 e x , independently of t x 0, (113)

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THE POISSON PROCESS AND RELATED PROCESSES

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Proof Fix t 0 The event { t > x} occurs only if one of the mutually exclusive events {X1 > t + x}, {X1 t, X1 + X2 > t + x}, {X1 + X2 t, X1 + X2 + X3 > t + x}, occurs This gives

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P { t > x} = P {X1 > t + x} +

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P {Sn t, Sn+1 > t + x}

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By conditioning on Sn , we nd P {Sn t, Sn+1 > t + x} = =

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P {Sn+1 > t + x | Sn = y} n P {Xn+1 > t + x y} n

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y n 1 y e dy (n 1)!

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y n 1 y e dy (n 1)!

This gives P { t > x} = e

(t+x)

n=1 0 t 0

e (t+x y) n

y n 1 y e dy (n 1)!

= e (t+x) +

e (t+x y) dy

= e (t+x) + e (t+x) (e t 1) = e x , proving the desired result The interchange of the sum and the integral in the second equality is justi ed by the non-negativity of the terms involved The theorem states that at each point in time the waiting time until the next arrival has the same exponential distribution as the original interarrival time, regardless of how long ago the last arrival occurred The Poisson process is the only renewal process having this memoryless property How much time is elapsed since the last arrival gives no information about how long to wait until the next arrival This remarkable property does not hold for general arrival processes (eg consider the case of constant interarrival times) The lack of memory of the Poisson process explains the mathematical tractability of the process In speci c applications the analysis does not require a state variable keeping track of the time elapsed since the last arrival The memoryless property of the Poisson process is of course closely related to the lack of memory of the exponential distribution Theorem 111 states that the number of arrivals in the time interval (0, s) is Poisson distributed with mean s More generally, the number of arrivals in any time interval of length s has a Poisson distribution with mean s That is, P {N (u + s) N (u) = k} = e s ( s)k , k! k = 0, 1, , (114)

independently of u To prove this result, note that by Theorem 112 the time elapsed between a given epoch u and the epoch of the rst arrival after u has the