final_practice_2009

To start nd the marginal densities of x and y px x py

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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 2-dimensional 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.

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