3.26 Space-time Poisson process Consider a highway that starts at x = 0 and extends infinitely eastward toward increasing values of x. Automobile accidents and breakdowns occur along the highway in a Poisson manner in time and space at a rate y per hour per mile. Any accident or breakdown that occurs remains at the location of occurrence until serviced.     At time t = 0, when there are no unserviced accidents or breakdowns on the highway, a helicopter starts from     x = 0 flying eastward above the highway at a constant speed s. As a service unit, the helicopter will land at the site of any accident or breakdown that it flies over. Moreover, given at time t the helicopter is located at x = st, the helicopter can be dispatched (by radio) to service any accident or breakdown that occurs behind it (i.e., at values of x st). We assume that any such dispatch occurs immediately after the accident or breakdown occurs.     We are interested in the time the helicopter first becomes busy, either by landing at an accident/breakdown site or by being dispatched to an accident/breakdown behind its current position; in the latter case, the instant of dispatch (not the time of arrival at the scene) is the time of interest.     Let T = time that the helicopter first becomes busy a. b. Let probability that the first accident/breakdown is a dispatch incident behind the helicopter 1 - probability that the first accident/breakdown occurs as a result of patrol (i.e., the helicopter     discovers it) Show that = 1 - = 1/2. Hint: Condition on the event that the first accident breakdown occurs in the time interval (t, t + dt). c. Let X   location of the first accident/breakdown that the helicopter services Show that d. Suppose that L1, = time of first accident/breakdown that the helicopter flies over, assuming that it is no longer dispatched by radio (i.e., all incidents are helicopter-discovered incidents) L2 = time of first accident/breakdown that the helicopter is dispatched to, assuming that it never services accident/breakdowns that it flies over Then, for instance, T = Min[L1, L2]. Show that L1 and L2 are identically distributed Rayleigh random variables, each with parameter sy. Finally, argue that L1 and L2 are independent, thereby concluding that the minimum of two independent Rayleigh random variables, each with parameter  y, is itself a Rayleigh random variable with parameter  2y.