Page 210 - Probability, Random Variables and Random Processes

RANDOM PROCESSES [CHAP 5

0 tl 5 Y c t,t t
II
H-- W(t) -I
cll-l
Fig. 5-14
Thus, using Eq. (5.1 64), we have
P[W(t)>z]=P(Z,>t-s+zIZ,>t-s)
- P(Zn>t-s+z) - e-a(t-s+r) - e-Lr
-
-
P(Zn>t-s) e-a(t-s) -
and P[W(t) 2 z] = 1 - ear (5.1 66)
which indicates that W(t) is an exponential r.v. with parameter I and is independent oft. Note that W(t) is
often called a waiting time.

5.56. Patients arrive at the doctor's office according to a Poisson process with rate 1 = & minute. The
doctor will not see a patient until at least three patients are in the waiting room.
Find the expected waiting time until the first patient is admitted to see the doctor.
What is the probability that nobody is admitted to see the doctor in the first hour?
Let T, denote the arrival time of the nth patient at the doctor's office. Then
T,=Z1 +z, +-+zn

where Z,, n = 1,2, . . . , are iid exponential r.v.'s with parameter I = &. By Eqs. (4.108) and (2.50),

The expected waiting time until the first patient is admitted to see the doctor is
E(T,) = 3(10) = 30 minutes
Let X(t) be the Poisson process with parameter I = &. The probability that nobody is admitted to see
the doctor in the first hour is the same as the probability that at most two patients arrive in the first 60
minutes. Thus, by Eq. (5.53,
P[X(60) - X(0) I 21 = P[X(60) - X(0) = 01 + P[X(60) - X(0) = 11 + P[X(60) - X(0) = 21
- e-60/10 + e-60/~o (rn) + e-60/10+(g)2
-
60
= e-6(1 + 6 + 18) x 0.062
T, denote the time of the nth event of a Poisson process X(t) with rate A. Suppose that one
event has occurred in the interval (0, t). Show that the conditional distribution of arrival time T,
is uniform over (0, t).
For z I t,

which indicates that T, is uniform over (0, t) [see Eq. (2.45)].