# hw7_sol - EE351K Probability Statistics and Random...

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Unformatted text preview: EE351K Probability, Statistics and Random Processes Spring 2009 Instructor: Vishwanath/Shakkottai { sriram, 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 (2 , 1 , 9 , 16 , . 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 ] = 2 + 1 = 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 ) = 9 + 16 + 2 * . 3 * 3 * 4 = 32 . 2 . 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 (32 . 2) e- 1 2 ( y- 3) 2 32 . 2 (b) Since ( W,Z ) is a linear tranformation of ( X 1 ,X 2 ) , these are jointly Gaussian as well. We can determine the parameters as follows. E [ W ] = 3 * E [ X 1 ] + E [ X 2 ] = 7 . Next, V ar ( W ) = 9 * V ar ( X 1 ) + V ar ( X 2 ) + 2 * 3 * Cov ( X 1 ,X 2 ) = 9 * 9 + 16 + 2 * 3 * . 3 * 3 * 4 = 118 . 6 . Similarly, E [ Z ] = 2- 1 = 1 and V ar ( Z ) = 9 + 16- 2 * . 3 * 3 * 4 = 17 . 8 . Finally, Cov ( W,Z ) = 3 * V ar ( X 1 )- 3 * Cov ( X 1 ,X 2 ) + Cov ( X 1 ,X 2 )- V ar ( X 2 ) = 3 , 8 . Hence, ρ WZ = 3 . 8 / √ 118 . 6 * 17 . 8 = 0 . 083 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 ) 2 E [ X 2 1 ] + 2 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 ]- 2 b (3- a ) E [ X 1 ] + E [ X 2 2 ]- 2 bE [ X 2 ] + b 2 = 13(3- a ) 2 + 2(3- a )(5 . 6)- b (3- a )(4) + 17- 2 b (1) + 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:- 26(3- a )- 11 . 2 + 4 b = 0 4( a- 3)- 2 + 2 b = 0 ....
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hw7_sol - EE351K Probability Statistics and Random...

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