Suppose in Example 5.19 that a car can overtake a slower moving car without any loss of speed. Suppose a car that enters the road at time s has a free travel time equal to to. Find the distribution of the total number of other cars that it encounters on the road (either by passing or by being passed).

Example 5.19 (A One Lane Road with No Overtaking) Consider a one lane road with a single entrance and a single exit point which are of distance L from each other (See Figure 5.2). Suppose that cars enter this road according to a Poisson process with rate, and that each entering car has an attached random value V which represents the velocity at which the car will travel, with the proviso that whenever the car encounters a slower moving car it must decrease its speed to that of the slower moving car. Let Vi denote the velocity value of the ith car to enter the road, and suppose that V, i > 1 are independent and identically distributed and, in addition, are independent of the counting process of cars entering the road. Assuming that the road is empty at time 0, we are interested in determining (a) the probability mass function of R(t), the number of cars on the road at time t; and (b) the distribution of the road traversal time of a car that enters the road at time y. Solution: Let T = L/V; denote the time it would take car i to travel the road if it were empty when car i arrived. Call T; the free travel time of car i, and note that T, T2,... are independent with distribution function G(x) = P(T x) = P(L/V x) = P(V > L/x) Let us say that an event occurs each time that a car enters the road. Also, lett be a fixed value, and say that an event that occurs at time s is a type 1 event if both a Figure 5.2 Cars enter at point a and depart at b. s < t and the free travel time of the car entering the road at time s exceeds t - s. In other words, a car entering the road is a type 1 event if the car would be on the road at time t even if the road were empty when it entered. Note that, independent of all that occurred prior to time s, an event occurring at times is a type 1 event with probability P(s) = JG (t t-s), if s t Letting N (y) denote the number of type 1 events that occur by time y, it follows from Proposition 5.3 that N (y) is, for y < t, a Poisson random variable with mean f*G(t-s)ds, y thus showing that P(R(t) = 0) = P(N (t) = 0) = e(t-s) ds To determine P(R(t) = n) for n > 0 we will condition on when the first type 1 event occurs. With X equal to the time of the first type 1 event (or to if there are no type 1 events), its distribution function is obtained by noting that X 0 Fx (y) = P(X < y) = P(N (y) > 0) = 1 - e-G(t-s) ds Differentiating gives the density function of X: fx (y) = (ty) e-Gt-s)ds, y 0 e-(t-y) ((t-y))"-1 (n-1)! Substituting this into Equation (5.19) yields - Se e- f G(u) du -(t-y) b ((t - y))n-1 (n 1)! if y