ch3 - Poisson Processes Poisson Distribution and the...

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1 Poisson Processes Chapter 3 2 Poisson Distribution and the Poisson Process Chapter 3.1 3 Poisson Distribution z The Poisson distribution with parameter µ > 0 is given by z Let X be Poisson r.v. Then, E[X]= µ . z Moreover, E[X 2 ] = µ 2 + µ and z Var(X) = µ . {} ! Pr k e p k X k k µ = = = 4 Theorem 3.1 z Let X and Y be independent random variables having Poisson distributions with parameters µ and ν . Then the sum X+Y has a Poisson distribution with parameter µ + ν .
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5 Proof {} { } { } ! ) ( ! )! ( ! Pr Pr , Pr Pr ) ( 0 ) ( 0 0 0 n e k n n e k n e k e k n Y k X k n Y k X n Y X n k n k n k k n n k k n k n k ν µ + = = = = = = = = = = + + = + = = = 6 Theorem 3.2 z Let N be a Poisson r.v. with parameter µ , and conditional on N, let M have a binomial distribution with parameters N and p. Then the unconditional distribution of M is Poisson with parameter µ p. 7 Proof { } { } [] ! ) ( ! ) ( )! ( ) 1 ( ! ) ( ! ) 1 ( )! ( ! ! Pr | Pr Pr 0 ) 1 ( 0 0 k p e e k e p k n p k e p n e p p k n k n n N n N k M k M k p n p k k n k n n k n k n = = = = = = = = = = = = 8 The Poisson process z A Poisson process of intensity or rate λ > 0 is an integer-valued stochastic process {X(t); t 0} for which For any time points t 0 =0< t 1 < … < t n , the process increments,X(t 1 )-X(t 0 ),…, X(t n )-X(t n-1 ), are independent r.v. For s 0 and t > 0, the r.v. X(s+t) – X(s) has the Poisson distribution And X(0) = 0. ,... 1 , 0 , ! ) ( ) ( ) ( Pr = = = + k k t e k s X t s X k t λ
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9 Example 3.1 z Defects occur along an undersea cable according to a Poisson process of rate λ = 0.1 per mile. z A) What is the probability that no defects appear in the first two miles of cable? z B) Given that there are no defects in the first two miles of cable, what is the conditional probability of no defect between points two and three? 10 Example 3.1 z To answer (A), observe that X(2) follows Poisson with parameter (0.1)(2) = 0.2. z Hence, Pr{X(2) = 0} = e -0.2 = 0.8187. z For part (B), we use the independence of X(3) – X(2) and X(2) – X(0) = X(2). Thus the conditional prob is the same as the unconditional prob. Pr{X(3) – X(2) = 0} = Pr{X(1) = 0} =e -0.1 = 0.9048. 11 Example 3.2 z Customers arrive in a certain store according to a Poisson process of rate λ =4 per hour. z Given that the store opens at 9:00 am, what is the probability that exactly one customer has arrived by 9:30 and a total of five has arrived by 11:30am? 12 Example 3.2 z Convert into probability language. We are asked to find Pr{X(1/2) =1,X(5/2) =5}. z We use the independence property: Pr{X(1/2) =1,X(5/2) =5} = Pr{X(1/2) =1,X(5/2) – X(1/2) =4} = e -4(1/2) 4(1/2){e -4(2) [4(2)] 4 /4!} =(2e -2 )[(512/3)e -8 ]=0.0154965.
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13 Nonhomogeneous Processes z Nonhomogeneous Poisson process has all the property of Poisson process except that the second property is replaced by z Where λ (t) is a function of time t.
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This note was uploaded on 05/21/2011 for the course STA 3007 taught by Professor Kb during the Spring '11 term at CUHK.

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ch3 - Poisson Processes Poisson Distribution and the...

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