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.