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Unformatted text preview: x + y  ≤ 1 &
x − y  ≤ 1
otherwise 1/2
0 1 pX,Y (x, y ) = (b) X 1 0 ≤ x, y ≤ 1
0 otherwise
(c) Solution:
(a) First we’ll ﬁnd the entries of the covariance matrix. To start, ﬁnd the marginal densities of X and Y :
pX (x) = pY (y ) = ∞
−∞
∞ pX,Y (x, y ) dy =
pX,Y (x, y ) dx = −∞ y =1−x
y =0
x=1−y
x=0 2 dy = 2(1 − x) for 0 ≤ x ≤ 1
2 dx = 2(1 − y ) for 0 ≤ y ≤ 1 We could have obtained the second result by symmetry. Now we can ﬁnd the moments:
E (X ) =
E (X 2 ) = ∞
−∞
∞ 1 xpX (x) dx =
0 σXX 0
2 = E (X ) − [E (X )] =
2 By symmetry we get:
E (Y ) = 1 x2 pX (x) dx = −∞ 2x(1 − x) dx = 1
3 2x2 (1 − x) dx = 1
6 1
18 1
1
1
, E (Y 2 ) = , σY Y =
3
6
18 Finally:
σXY = E [(X − mX )(Y − mY )] = 1
y =0 1−y
x=0 (xy )2 dx dy − mX mY =
1 6 1
y =0 y (1 − y )2 dy − 1
1
=−
9
36 6. Let Z be a 2dimensional Gaussian random vector with mean mZ and covariance matrix CZ as speciﬁed below:
Z= W
V mZ = , 0
0 , CZ = 21
14 . 2
Let X be a Gaussian random variables with mean mX = 2 and variance σX = 8. Assume X and Z are statistically
independent. The random variable Y is related to X and Z as follows: Y = (2 + W )X + V
(a) Find the linear minimum mean squared error estimate of X based on Y .
(b) Determine the resulting mean squared error. 7...
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This note was uploaded on 01/08/2013 for the course ECE 531 taught by Professor Natasha during the Spring '10 term at Ill. Chicago.
 Spring '10
 Natasha

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