3.26 Spacetime
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
L_{1}, =
time of first
accident/breakdown that the helicopter flies over, assuming that
it is no longer
dispatched by radio (i.e., all incidents are
helicopterdiscovered incidents)
L_{2} = 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[L_{1},
L_{2}].
Show that L_{1} and
L_{2}
are identically distributed Rayleigh random variables, each with
parameter sy.
Finally, argue that L_{1}
and L_{2}
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.
