Conditioned on the latter information the arrival

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Conditioned on the latter information, the arrival time is uniformly distributed in the interval [ k, k + 1] (cf. Problem 6.18), which implies that X isuniformly distributed in [0 , 1]. Since this conditional distribution of X is the same for every k , it follows that X is independent of K . Let D be the departure time of the job of interest. Since the job stays in the system for an integer amount of time, we have that D is of the form D = L + X , where L is a nonnegative integer. Since the job stays in the system for a geometrically distributed amount of time, and the geometric distribution has the memorylessness property, it follows that L is also memoryless. In particular, L is similar to a geometric random variable, except that its PMF starts at zero. Furthermore, L is independent of X , since X is determined by the arrival process, whereas the amount of time a job stays in the system is independent of the arrival process. Thus, D is the sum of two independent random variables, one uniform and one geometric. Therefore, D has “geometric staircase” PDF, given by f D ( d ) = 1 2 d , d 0 , and where d stands for the largest integer below d . Solution to Problem 6.16. (a) The random variable N is equal to the number of successive interarrival intervals that are smaller than τ . Interarrival intervals are independent and each one is smaller than τ with probability 1 e λτ . Therefore, P ( N = 0) = e λτ , P ( N = 1) = e λτ ( 1 e λτ ) , P ( N = k ) = e λτ ( 1 e λτ ) k , so that N has a distribution similar to a geometric one, with parameter p = e λτ , except that it shifted one place to the left, so that it starts out at 0. Hence, E [ N ] = 1 p 1 = e λτ 1 . (b) Let T n be the n th interarrival time. The event { N n } indicates that the time between cars n 1 and n is less than or equal to τ , and therefore E [ T n | N n ] = E [ T n | T n τ ]. Note that the conditional PDF of T n is the same as the unconditional one, except that it is now restricted to the interval [0 , τ ], and that it has to be suitably renormalized so that it integrates to 1. Therefore, the desired conditional expectation is E [ T n | T n τ ] = τ 0 sλe λs ds τ 0 λe λs ds . This integral can be evaluated by parts. We will provide, however, an alternative approach that avoids integration. We use the total expectation formula E [ T n ] = E [ T n | T n τ ] P ( T n τ ) + E [ T n | T n > τ ] P ( T n > τ ) . 76
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We have E [ T n ] = 1 , P ( T n τ ) = 1 e λτ , P ( T n > τ ) = e λτ , and E [ T n | T n > τ ] = τ + (1 ). (The last equality follows from the memorylessness of the exponential PDF.) Using these equalities, we obtain 1 λ = E [ T n | T n τ ] ( 1 e λτ ) + τ + 1 λ e λτ , which yields E [ T n | T n τ ] = 1 λ τ + 1 λ e λτ 1 e λτ . (c) Let T be the time until the U-turn. Note that T = T 1 + · · · + T N + τ . Let v denote the value of E [ T n | T n τ ]. We find E [ T ] using the total expectation theorem: E [ T ] = τ + n =0 P ( N = n ) E [ T 1 + · · · + T N | N = n ] = τ + n =0 P ( N = n ) n i =1 E [ T i | T 1 τ, . . . , T n τ, T n +1 > τ ] = τ + n =0 P ( N = n ) n i =1 E [ T i | T i τ ] = τ + n =0 P ( N = n ) nv = τ + v E [ N ] , where E [ N ] was found in part (a) and v was found in part (b). The second equality
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