Page 208 - Probability, Random Variables and Random Processes

RANDOM PROCESSES [CHAP 5

Substituting the above expression into Eq. (5.156), we have

Integrating, we get

Since pn(0) = 0, c, = 0, and we obtain

which is Eq. (5.55) of Definition 5.6.1. Thus we conclude that Definition 5.6.2 implies Definition 5.6.1.

5.50. Verify Eq. (5.59).
We note first that X(t) can assume only nonnegative integer values; therefore, the same is true for the
counting increment X(t + At) - X(t). Thus, summing over all possible values of the increment, we get
00
1 P[X(t + At) - X(t) = k] = P[X(t + At) - X(t) = 01
k = 0
+ P[X(t + At) - X(t) = 11 + P[X(t + At) - X(t) 2 21
= 1
Substituting conditions 3 and 4 of Definition 5.6.2 into the above equation, we obtain
P[X(t + At) - X(t) = 0) = 1 - A At + o(At)

5.51. (a) Using the Poison probability distribution in Eq. (5.158), obtain an analytical expression for
the correction term o(At) in the expression (condition 3 of Definition 5.6.2)
P[X(t + At) - X(t) = 11 = A At + o(At) (5.1 59)
(b) Show that this correction term does have the property of Eq. (5.58); that is,

o(At)
lim - - 0
-
at-o At
(a) Since the Poisson process X(t) has stationary increments, Eq. (5.159) can be rewritten as
P[X(At) = 11 = p,(At) = A At + o(At) (5.1 60)
Using Eq. (5.1 58) [or Eq. (5.1 57)], we have
pl(At) = L At e-at = A At(1 + - 1)
= L At + A At(eVAAt - 1)
Equating the above expression with Eq. (5.160), we get

from which we obtain
o(At) = A At(e-"t - 1)
(b) From Eq. (5.161), we have

- lim A AWUt - 1) = lim *((?-at - 1) = 0
-
lim -
At-0 At At-0 At At-0
5.52. Find the autocorrelation function R,(t, s) and the autocovariance function K,(t, s) of a Poisson
process X(t) with rate 1.