20112ee131B_1_EE131B_hw1_sol

20112ee131B_1_EE131B_hw1_sol - EE131B HW1 1 Problem 2 The...

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Unformatted text preview: EE131B HW1 1 Problem 2 The Gaussian random variable x with zero mean and variance 2 has the fol- lowing density function: p ( x ) = 1 2 e- x 2 2 2 Thus the corresponding chharacteristic function of this density is C ( t ) = + Z- e itx 1 2 e- x 2 2 2 dx = + Z- 1 2 e- ( x- i 2 t ) 2 + 4 t 2 2 2 dx = e- 2 2 t 2 + Z- 1 2 e- ( x- i 2 t ) 2 2 2 dx = e- 2 2 t 2 With Taylor expansion, we have C ( t ) = e- 2 2 t 2 = X n =0 (- 2 t 2 2 ) n n ! = 1- 2 2 t 2 + 4 8 t 4- ... Therefore, we denote n = k/ 2 when k is even, and we can conclude that the k th moment of this Gaussian random variable is m k = k is odd k ! k 2 k 2 ( k 2 )! k is even 1 2 Problem 6 (a) C ( t ) = + Z- e itx p ( x ) dx = +1 / 2 Z- 1 / 2 e itx dx = 1 it e itx | +1 / 2- 1 / 2 = e it/ 2- e- it/ 2 it = 2 i sin t 2 it = sin t 2 t 2 (b) With the inverse transform of characteristic function, p ( x ) =...
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This note was uploaded on 02/20/2012 for the course EE 131B taught by Professor Balakrishnan during the Spring '11 term at UCLA.

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20112ee131B_1_EE131B_hw1_sol - EE131B HW1 1 Problem 2 The...

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