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hw7_sol - EE351K Probability Statistics and Random...

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EE351K Probability, Statistics and Random Processes Spring 2008 Instructor: Shakkottai { shakkott } @ece.utexas.edu Homework 7: Solutions Problem 1 Let ( X 1 , X 2 ) be jointly Gaussian random variables with ( X 1 , X 2 ) N (1 , 2 , 4 , 9 , 0 . 3) . a) Determine the pdf of Y = X 1 + X 2 . b) Determine the joint pdf of ( W, Z ) where W = 3 X 1 + X 2 and Z = X 1 - X 2 . c) Determine an expression for E [ W | X 1 ] and E [ W | Z ] . Solution: Since X 1 and X 2 are jointly Gaussian, any linear combination is a Gaussian random variable. (a) Y = X 1 + X 2 is Gaussian. To determine the parameters, E [ Y ] = E [ X 1 + X 2 ] = 1 + 2 = 3 . Also, V ar ( Y ) = V ar ( X 1 ) + V ar ( X 2 ) + 2 Cov ( X 1 , X 2 ) . Thus, we have V ar ( Y ) = 4 + 9 + 2 * 0 . 3 * 2 * 3 = 16 . 6 . Hence, pdf of Y is: f Y ( y ) = 1 q 2 πσ 2 y e - 1 2 ( y - μy ) 2 σ 2 y = 1 p 2 π (16 . 6) e - 1 2 ( y - 3) 2 16 . 6 (b) Since ( W, Z ) is a linear tranformation of ( X 1 , X 2 ) , these are jointly Gaussian as well. Similar to the calculation in class, we can determine the parameters as follows. E [ W ] = 3 * E [ X 1 ] + E [ X 2 ] = 5 . Next, V ar ( W ) = 9 * V ar ( X 1 ) + V ar ( X 2 ) + 2 * 3 * Cov ( X 1 , X 2 ) = 9 * 4 + 9 + 2 * 3 * 0 . 3 * 2 * 3 = 55 . 8 . Similarly, E [ Z ] = 1 - 2 = - 1 and V ar ( Z ) = 4 + 9 - 2 * 0 . 3 * 2 * 3 = 9 . 4 . Finally, Cov ( W, Z ) = 3 * V ar ( X 1 ) - 3 * Cov ( X 1 , X 2 ) + Cov ( X 1 , X 2 ) - V ar ( X 2 ) = - 4 . 2 . Hence, ρ WZ = - 4 . 2 / 55 . 8 * 9 . 4 = - 0 . 18 . c) Since for a Gaussian, the MMSE is the linear estimator, we can write: E [ W | X 1 ] = aX 1 + b , where a and b minimize E [( W - aX 1 - b ) 2 ] . E [(3 X 1 + X 2 - aX 1 - b ) 2 ] = (3 - a ) E [ X 2 1 ] + E [(3 - a ) X 1 ( X 2 - b )] + E [( X 2 - b ) 2 ] = (3 - a ) 2 E [ X 2 1 ] + (3 - a ) E [ X 1 X 2 ] - b (3 - a ) E [ X 1 ] + E [ X 2 2 ] - 2 bE [ X 2 ] + b 2 = 5(3 - a ) 2 + (3 - a )(3 . 8) - b (3 - a )(1) + 13 - 2 b (2) + b 2 . Denote this quantity by f ( a, b ) . To find the linear estimator, we use: ∂f ( a,b ) ∂a = 0 and ∂f ( a,b ) ∂b = 0 . So we need to solve the following system of two equations with two unknowns: - 10(3 - a ) - 3 . 8 + b = 0 ( a - 3) - 4 + 2 b = 0 . Get a = 3 . 189 and b = 1 . 9 , so E [ W | X 1 ] = 3 . 189 X 1 + 1 . 9 To find E [ W | Z ] , we use the same procedure, where we minimize E [( W - cZ - d ) 2 ] . E [( W - cZ - d ) 2 ] = E [(3 X 1 + X 2 - c ( X 1 - X 2 ) - d ) 2 ] = (3 - c ) 2 E [ X 2 1 ] + (3 - c ) E [ X 1 ((1 + c ) X 2 - d )]+ E [((1+ c ) X 2 - d ) 2 ] = (3 - c ) 2 (5)+(3 - c )(1+ c )(3 . 8) - d (3 - c )+(1+ c ) 2 (13)+ d 2 - 2 d (1+ c )(2) . Again we denote the result by g(c,d) and we use: ∂g ( c,d ) ∂c = 0 and ∂g ( c,d ) ∂d = 0 .
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So we need to solve the following system of two equations with two unknowns: - 10(3 - c ) + 3 . 8( - 2 c + 2) + d + 13(1 + c )(2) - 4 d = 0 ( c - 3) + 2 d - 4(1 + c ) = 0 .
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