Page 203 - Schaum's Outlines - Probability, Random Variables And Random Processes

196 RANDOM PROCESSES [CHAP 5

Now by the result of Prob. 5.39, we see that absorption is certain; therefore '

Thus Y(k) = pY(k + 1) + qY(k - 1) + p + q

Rewriting Eq. (5.1 do), we have

Thus, finding P(k) reduces to solving Eq. (5.141) subject to the boundary conditions given by Eq. (5.137).
Let the general solution of Eq. (5.141) be

where &(k) is the homogeneous solution satisfying

and Y,(k) is the particular solution satisfying
1 4 1
Yp(k + 1) - - Yp(k) + - Y,(k - 1) = - -
P P P
Let Y,(k) = ak, where a is a constant. Then Eq. (5.143) becomes
1 4 1
(k + 1)a - - ka + - (k - 1)a = - -
P P P
from which we get a = l/(q - p) and
k
Y,(k) = - P Z 4
4-P
Since Eq. (5.142) is the same as Eq. (5.131), by Eq. (5.133), we obtain

where c, and c2 are arbitrary constants. Hence, the general solution of Eq. (5.141) is

Now, by Eq. (5.137),
Y(0) = 0-q + c, =o

Solving for c, and c,, we obtain

Substituting these values in Eq. (5.146), we obtain (for p # q)

When p = q = 4, we have
Y(k) = E(T,) = k(N - k) p = q = 4