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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 2 y e 1 2 ( y y ) 2 2 y = 1 p 2 (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, 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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 Spring '07
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