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Unformatted text preview: EE 351K Probability, Statistics, and Random Processes SPRING 2003
Instructor: Sanjay Shakkottai
Random Processes and Stationarity —————————_—__ Probleml Let {1’36 are integers Le. a' can be 0, positive, and negative) be a sequence of i.i.d.
Bernoulli random variables with parameter 1), Le. 12={G w.p. 1—1;:
1 w.p. p Let s{t) be the standard pulse function. In other words, 0 t<0,
U 3}]. Let us deﬁne the random process (for all times) in the following way.
DO
Xe) = Z nee — a)
.w'=—oo a) Prove that X (t) is not wide sense stationary. b) Let R be a uniform random variable over the interval [0, 1]. We deﬁne the random process Z (t) (for all
times) in the following manner. 20!): ‘Z lass—j—R) J=—oo Prove that Z (t) is a widesense stationary process. Solution One realization of the random process X (t) can be seen in Figure l. t1ﬁ1+T [bigi"? Figure l: A Realization of X0) For example, the value in the interval [2, 3] is determined by the random variable Y2 since 50‘. ~ 2) = 1 only
if 2 g t g 3. Compare E[X(:1)X(r1 + 1—)] with E[X(t'2_)X(tg + r)]. E[X(t1)X(t1 + 7)] = Bum/1] = E[%]E[Y1] = p2 and E[X(t2)X(t2 + T)] = E[Y31’3] = E[Y32] = ,7). Notice that the autocorrelations of
two cases are different even though we choose same time difference (7'). Therefore, X (t) is not widesense
stationary. On the other hand, Z ( t) is widesense stationary. Notice that Z (t) is randomtime shifted version
of X(t). When ‘7' is greater than 1, E[X(t)X(t + 7)] = E[}’;}’}] = p2(i 96 j) meaning the autocorrelation
is not dependent upon the choice of t. Lets think about the case when 1" is smaller than 1. E[X(t)X(t + 7)] = E[Y_11ﬁ]P{R < T) + E[Y_1Y_1]P(R > 7‘)
'2 E[Y_1]E[}’ﬁ]r + E[Y31](1 — 7') = 1027' +p(1 — 7') Therefore, the result is a function of only T , not the choice of 2?. Problem 2 Let us deﬁne a sequence oi'i.i.d. random variables {X1, X2, . . .}, with X, ~ Uniform[—1,1], 2': 1,2,3,. .. Deﬁne,
1’; 2 X3, j=1,2,3,... (1) H 3'
1 .
2, 72m, 321,2,3,... (2)
3 i=1
For each of the four parts, justify your answer. (a). Isthe random process {X,,i = 1, 2, 3, . . .} widesense stationary?
(b). Is the random process {1’},j = 1, 2, 3, . . .} widesense stationary? (c). Is the random process {2333‘ = l, 2, 3, . . } wide~sense stationary? Solution a) Yes. It is clear that E[X,] = (LW and E[X,X,] = E[X,~]E[Xj] = 0 ift' 75 3'. Otherwise, ifi = j,
E[X,Xj] = E[Xf] = 1/2. Therefore, the autocorrelation E[X,X_,] does not depend on i. Hence, the
random process {X,,i = l, 2, 3, . . .} is wide~sense stationary. b) Yes. EH3] = E[X3] = U,Vj. Also, EDGEj = EfXng] = E[X§] = 0.5. That is, the autocorrelation
E[}’;}f,] 1s a constant and does not depend on '3'. Hence, the random process {YEJ = 1, 2, 3, . . .} is wide— c) No. This is because, 1 r 1 J"+k
EiZjZki = Ekigxdmﬁgmj H
U}
Tﬁ'
C“
+ H
E:
M
L;
35;.
_.____,
+
ti}
to.
C?
+ H
3.:
Me;
ME
3*:
35
I____. i: 1=1i=1,£}‘£t
1
= , EX? +0
1 l (i Hence, the autocorrclation, E[Zij+k] depends on 3‘. Thus, the random process {Zj,j = 1, 2,3, . . .} is
widesense stationary. ...
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