Page 183 - Schaum's Outlines - Probability, Random Variables And Random Processes

176 RANDOM PROCESSES [CHAP 5

5.7. Consider the random process X(t) of Prob. 5.4. Determine the pdf s of X(t) at t = 0, n/4w, 42w,
n/w.

For t = 0, X(0) = Y cos 0 = Y. Thus,

For t = rr/4o, X(n/4o) = Y cos n/4 = 1/$ Y. Thus,
O<x<l/JZ
otherwise
For t = 420, X(zl2o) = Y cos n/2 = 0; that is, X(n/20) = 0 irrespective of the value of Y. Thus, the
pmf of X(o/2o) is

1 -l<x<O
0 otherwise

5.8. Derive the first-order probability distribution of the simple random walk X(n) of Prob. 5.2.

The first-order probability distribution of the simple random walk X(n) is given by

where k is an integer. Note that P(Xo = 0) = 1. We note that p,(k) = 0 if n < 1 k 1 because the simple random
walk cannot get to level k in less than I k I steps. Thus, n 2 1 k I.
Let Nnf and N,- be the r.v.'s denoting the numbers of + 1s and - Is, respectively, in the first n steps.
Then

Adding Eqs. (5.68) and (5.69), we get
Nnt = $(n + X,) (5.70)
Thus, X, = k if and only if N,+ = i(n + k). From Eq. (5.70), we note that 2N,+ = n + X, must be even.
Thus, X, must be even if n is even, and X, must be odd if n is odd. We note that N,+ is a binomial r.v. with
parameters (n, p). Thus, by Eq. (2.36), we obtain

where n 2 ( k 1, and n and k are either both even or both odd.

5.9. Consider the simple random walk X(n) of Prob. 5.2.

(a) Find the probability that X(n) = - 2 after four steps.
(b) Verify the result of part (a) by enumerating all possible sample sequences that lead to the
value X(n) = - 2 after four steps.
(a) Setting k = -2 and n = 4 in Eq. (5.71), we obtain