Page 206 - Probability, Random Variables and Random Processes

RANDOM PROCESSES [CHAP 5

(c) Setting a = a and b = in Eq. (5.120) (Prob. 5.30), we get

Since limn,, (b)" = 0, we obtain

lim Pn = lim
n-co

POISSON PROCESSES
5.48. Let T, denote the arrival time of the nth customer at a service station. Let 2, denote the time
interval between the arrival of the nth customer and the (n - 1)st customer; that is,
Z,=T,-T,-, nrl (5.149)

and To = 0. Let (X(t), t 2 0) be the counting process associated with {T,, n 2 0). Show that if
X(t) has stationary increments, then Z, , n = 1,2, . . . , are identically distributed r.v.3.
We have

By Eq. (5.149), P(Z,>z)= P(T,- T,-, > z)= P(T,> T,-, +z)
Suppose that the observed value of T,-, is t,- ,. The event (T, > T,-, + z ( T,-, = tn- ,) occurs if and only if
X(t) does not change count during the time interval (tn- ,, tn-, + z) (Fig. 5-13). Thus,

Since X(t) has stationary increments, the probability on the right-hand side of Eq. (5.150) is a function only
of the time difference z. Thus

which shows that the conditional distribution function on the left-hand side of Eq. (5.1 51) is independent of
the particular value of n in this case, and hence we have
F,,(z) = P(Zn < Z) = 1 - P[X(Z) = 01 (5.1 52)
which shows that the cdf of Zn is independent of n. Thus we conclude that the 2,'s are identically distrib-
uted r.v.'s.

0 5- 5 t
l (2 1
Fig. 5-13

5.49. Show that Definition 5.6.2 implies Definition 5.6.1.
Let pn(t) = P[X(t) = n]. Then, by condition 2 of Definition 5.6.2, we have
p,(t + At) = P[X(t + At) = 01 = P[X(t) = 0, X(t + At) - X(0) = 01
= P[X(t) = 01 P[X(t + At) - X(t) = 01