Homework5_Solution

Homework5_Solution - 1 Suppose that people arrive at a bus...

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1. Suppose that people arrive at a bus stop in accordance with a Poisson process with rate λ . The bus departs at time t . Let X denote the total amount of waiting time of the person who ﬁrst arrives at the bus stop. Let Y denote the total amount of waiting time of all those who get on the bus at time t . a) Find E [ X ]. b) Find E [ Y ]. c) Find V ar ( Y ). Solution to Problem 1: The number of passengers who get on the bus at time t is given by N ( t ). If N ( t ) = 0, no customers arrive before the bus departs and we deﬁne the waiting time to be zero. a) Method 1: Given N ( t ) = n > 0, S 1 has the same distribution as U (1) , the smallest of U 1 ,...,U n , where U 1 ,U 2 , ··· ,U n are independent uniform (0 ,t ) random variables. E [ X | N ( t ) = n ] = E [ t - U (1) ] . Since P ( U (1) > s ) = P ( U i > s for all i ) = ± t - s t ² n , we have the probability density function of U (1) , f U (1) ( s ) = n ( t - s ) n - 1 /t n . Consequently E [ t - U (1) ] = Z t 0 ( t - s ) n ( t - s ) n - 1 t n ds = nt n + 1 . So E [ X ] = X n =0 nt n + 1 e - λt ( λt ) n n ! = te - λt ( X n =0 ( λt ) n n ! - 1 λt X n =0 ( λt ) n +1 ( n + 1)! ) = te - λt ( e λt - 1 λt ( e λt - 1)) = t (1 - 1 λt (1 - e - λt )) = t - 1 λ (1 - e - λt ) . Method 2: Let T 1 be the arrival time of the ﬁrst passenger. We know that T 1 follows exponential distribution with parameter λ . Then we have X = ³ t - T 1 T 1 < t 0 otherwise 1

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Then E [ X ] = E [ t - T 1 | T 1 < t ] P ( T 1 < t ) + 0 = t · P ( T 1 < t ) - E [ T 1 | T 1 < t ] P ( T 1 < t ) = t (1 - e - λt ) - Z t 0 sλe - λs ds = t - 1 λ (1 - e - λt ) . b) Since, given
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Homework5_Solution - 1 Suppose that people arrive at a bus...

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