Page 213 - Probability, Random Variables and Random Processes

CHAP. 51 RANDOM PROCESSES

Let X =

Then Eq. (5.1 78) can be expressed as

Y=AX

Then the characteristic function for Y can be written as

Since X is a normal random vector, by Eq. (5.1 77) we can write
Yx(ATm) = e~p[j(A~m)~p~ - 3(AT~)TKX(ATw)]
= exp[ joTAp, - $wTAKx AT4
Thus Yda,, . . . , a,) = exp(jwTpy - $mTKy a)
where = K~ = AK~
Comparing Eqs. (5.1 77) and (5.180), we see that Eq. (5.180) is the characteristic function of a random vector
Y. Hence, we conclude that Y,, . . . , Ym are also jointly normal r.v.'s
Note that on the basis of the above result, we can say that a random process {X(t), t E T) is a normal
process if every finite linear combination of the r.v.'s X(ti), ti E T is normally distributed.

5.61. Show that a Wiener process X(t) is a normal process.
Consider an arbitrary linear combination

where 0 5 t, < - - - < tn and ai are real constants. Now we write
n
aiX(ti) = (a, + . . . + a,)[X(tl) - X(O)] + (a, + . + a,)[X(t,) - X(tl)]
i= 1

Now from conditions 1 and 2 of Definition 5.7.1, the right-hand side of Eq. (5.183) is a linear combination
of independent normal r.v.3. Thus, based on the result of Prob. 5.60, the left-hand side of Eq. (5.183) is also
a normal r.v.; that is, every finite linear combination of the r.v.'s X(ti) is a normal r.v. Thus we conclude that
the Wiener process X(t) is a normal process.

5.62. A random process {X(t), t E T) is .said to be continuous in probability if for every 8 > 0 and t E T,
lim P( ( X(t + h) - X(t) I > E) = 0
h+O
Show that a Wiener process X(t) is continuous in probability.
From Chebyshev inequality (2.97), we have
Var[X(t + h) - X(t)]
P( I X(t + h) - X(t) I > E} I E > 0
&2