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University of Illinois Fall 1998
ECE 313: Solutions to Problem Set
- The easiest way to solve this problem is the indirect way. For each i = 1 ; 2 ; : : : ; n, we de ne the random variables:
Xi =
1 i-th die shows 1 0 otherwise
Yi =
1 i-th die shows 2 0 otherwise
Since the same die cannot land on b oth 1 and 2, we conclude that Xi Yi = 0 always. This is a deterministic constant, so that E [Xi Yi ] = 0 for all i. It is furthermore easy to see that
E [Xi ] = P fXi = 1 g = 1 = 6 and E [Yi ] = P fYi = 1 g = 1 =6, for all i. Thus we have:
Cov (Xi ; Yj ) = E [Xi Yj ] E [Xi ]E [Yj ] =
if j = i
0 if j 6 = i
The second case follows from the fact that the n rolls of the dice are indep endent exp eriments,
and therefore Xi and Yj are indep endent random variables for j 6 = i. Now observe that
X = X 1 + X 2 + + Xn and Y = Y 1 + Y 2 + + Yn. Therefore, we obtain:
Cov (X ; Y ) = Cov
Xn
i=
Xi ;
X^ n
j =
Yj
A =
X^ n
i=
X^ n
j =
Cov (Xi ; Yj ) =
X^ n
i=
Cov (Xi ; Yi ) =
n 36
where the second equality follows from the bilinearity of the covariance function, the third equality follows from the fact that Cov (Xi ; Yj ) = 0 unless j = i, and the nal equality follows
from the fact that Cov (Xi ; Yi ) = 1 = 36 for all i.
2. (a) Since Z = 2(X + Y )(X Y ) = 2 X 2 2 Y 2 , we have
E [Z ] = 2(E [X 2 ] E [Y 2 ]) = 2(Var(X ) Var(Y ) + E 2 [X ] + E 2 [Y ]) = 40
(b) We have
Cov (T ; U ) = Cov (2X + Y ; 2 X Y )
= 4 Cov (X ; X ) 2 Cov (X ; Y ) + 2 Cov (Y ; X ) Cov (Y ; Y )
= 4 Var (X ) Var(Y ) = 7
(c) Here
E [W ] = E [3X + Y + 2] = 3 E [X ] + E [Y ] + 2 = 9
Var(W ) = Var(3X + Y ) = 9 Var (X ) + Var(Y ) + 6 Cov (X ; Y ) = 48 : 6
(d) If X and Y are jointly Gaussian, then their linear combination W = 3 X + Y + 2 is a Gaussian random variable. Therefore
P fW > 0 g = 1
@ 0 q^ ^ E^ [W^ ]
Var(W )
A = 1 p^0 ^9
p
- The key p oint to observe here is that X 1 ; X 2 ; : : : are indep endent random variables, and
therefore Yn and Yn+k are also indep endent if k 3. Thus Cov (Yn ; Yn+k ) = 0 if k 3. Now,
for k = 0, we have:
Cov (Yn ; Yn ) = Var (Yn ) = Var(Xn + Xn+1 + Xn+2 ) = Var(Xn ) + Var(Xn+1 ) + Var(Xn+2 ) = 3 2
where we have used the indep endence of Xn ; Xn+1 ; Xn+2 once again. For k = 1, we have:
Cov (Yn ; Yn+1 ) = Cov (Xn + Xn+1 + Xn+2 ; Xn+1 + Xn+2 + Xn+3 ) = Cov (Xn ; Xn+1 + Xn+2 + Xn+3 ) + Cov (Xn+1 + Xn+2 ; Xn+1 + Xn+2 + Xn+3 ) = 0 + Cov (Xn+1 + Xn+2 ; Xn+1 + Xn+2 ) + Cov (Xn+1 + Xn+2 ; Xn+3 ) = Var(Xn+1 + Xn+2 ) + 0 = Var (Xn+1 ) + Var(Xn+2 ) = 2 2
where we have used the bilinearity of the covariance function, along with the indep endence of Xn ; Xn+1 ; Xn+2 ; Xn+3. Finally, for k = 2, we have:
Cov (Yn ; Yn+2 ) = Cov (Xn + Xn+1 + Xn+2 ; Xn+2 + Xn+3 + Xn+4 ) = Cov (Xn + Xn+1 ; Xn+2 + Xn+3 + Xn+4 ) + Cov (Xn+2 ; Xn+2 + Xn+3 + Xn+4 ) = 0 + Cov (Xn+2 ; Xn+2 ) + Cov (Xn+2 ; Xn+3 + Xn+4 ) = Var(Xn+2 ) + 0 = 2
where the reasoning is similar to that used for the case k = 1. Summarizing all of the ab ove,
we conclude that Cov (Yn ; Yn+k ) = (3 k ) 2 for k = 0 ; 1 ; 2 and Cov (Yn ; Yn+k ) = 0 for k 3.
- De ne a = cos and b = sin . Thus a and b are real constants, and we have:
Z = aX + bY
W = bX + aY
or
Z W
a b
b a
X Y
Since this is a linear transformation of jointly Gaussian random variables, we can conclude that Z and W are joinly Gaussian. Therefore, Z and W are indep endent if and only if they are uncorrelated. Thus, we need to solve
Cov (Z ; W ) = Cov (aX + bY ; bX + aY )
= abCov (X ; X ) + a^2 Cov (X ; Y ) b^2 Cov (Y ; X ) + abCov (Y ; Y )
= ab
Var(Y ) Var(X )
+ (a^2 b^2 )Cov (X ; Y )
= cos sin ( 22 21 ) + (cos^2 sin^2 ) 1 2
= 0 : 5 sin 2 ( 22 12 ) + cos 2 1 2 = 0
for the value of . The solution tan 2 =
sp eci es the desired angle . Note that for 1 = 2 , we get = =4.
- (a) We have
= E [X ] =
Z 1
2
u 24 u ^4 du = 24
Z 1
2
u ^3 du = 3
Thus
q ( ) = P fjX 3 j > g = 1 P fjX 3 j < g
Z 3+
min(2; 3 )
24 u ^4 du =
1 (3 ^8 ) 3 + (3+^8 ) 3 < 1
8
(3+ )^3 ^ ^1
