In: Statistics and Probability
Assume that the Poisson process X = {X(t) : t ≥ 0} describes students’ arrivals at the library with intensity λ = 4 per hour. Given that the tenth student arrived exactly at the end of fourth hour, or W10 = 4, find:
1. E [W1|W10 = 4]
2. E [W9 − W1|W10 = 4].
Hint: Suppose that X {X(t) : t ≥ 0} is a Poisson process with rate λ > 0 and its arrival times are defined for any natural k as Wk = min[t ≥ 0 : X(t) = k] (1) Then for any natural m, the inter-arrival times, {T1 = W1, T2 = W2 − W1, . . . , Tm = Wm − Wm−1} are independent variables with the common exponential distribution, fT(t) = λ · e −λ·t for t > 0.