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Unformatted text preview: 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 , . 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 * . 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 * . 3 * 2 * 3 = 55 . 8 . Similarly, E [ Z ] = 1 2 = 1 and V ar ( Z ) = 4 + 9 2 * . 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 = . 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 ....
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This note was uploaded on 06/10/2008 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas at Austin.
 Spring '07
 BARD

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