Page 207 - Probability, Random Variables and Random Processes

CHAP. 51 RANDOM PROCESSES

Now, by Eq. (5.59), we have
P[X(t + At) - X(t) = 01 = 1 - r3. At + o(At)
Thus, po(t + At) = po(t)[l - I At + o(At)]

Letting At + 0, and by Eq. (5.58), we obtain

~b(t) - IP&)
=
Solving the above differential equation, we get
po(t) = ke-"
where k is an integration constant. Since po(0) = P[X(O) = 01 = 1, we obtain
po(t) = e - At
Similarly, for n > 0,
pn(t + At) = P[X(t + At) = n]
= P[X(t) = n, X(t + At) - X(0) = 01
+ P[X(t) = n - 1, X(t + At) - X(0) = 1) + n P[X(t) = n - k, X(t +At) - X(0) = k]
k=2
Now, by condition 4 of Definition 5.6.2, the last term in the above expression is o(At). Thus, by conditions 2
and 3 of Definition 5.6.2, we have
p,(t + At) = pn(t)[l - 1 At + o(At)] + p,- ,(t)[I At + o(At)] + @t)

Thus

and letting At -, 0 yields
=
PX~) + IP,~ Lpn - 1 (t)
Multiplying both sides by e", we get

d
Hence - [eapn(t)] = IeAtpn - ,(t)
dt
Then by Eq. (5.154), we have

or pl(t) = (At + ~)e-'~

where c is an integration constant. Since p,(O) = P[X(O) = 1) = 0, we obtain
p,(t) = Ate-*'
To show that

we use mathematical induction. Assume that it is true for n - 1 ; that is,