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20112ee131B_1_EE131B_hw1_sol

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

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EE131B HW1 1 Problem 2 The Gaussian random variable x with zero mean and variance σ 2 has the fol- lowing density function: p ( x ) = 1 2 πσ e - x 2 2 σ 2 Thus the corresponding chharacteristic function of this density is C ( t ) = + Z -∞ e itx 1 2 πσ e - x 2 2 σ 2 dx = + Z -∞ 1 2 πσ e - ( x - 2 t ) 2 + σ 4 t 2 2 σ 2 dx = e - σ 2 2 t 2 + Z -∞ 1 2 πσ e - ( x - 2 t ) 2 2 σ 2 dx = e - σ 2 2 t 2 With Taylor expansion, we have C ( t ) = e - σ 2 2 t 2 = X n =0 ( - σ 2 t 2 2 ) n n ! = 1 - σ 2 2 t 2 + σ 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 = 0 k is odd k ! σ k 2 k 2 ( k 2 )! k is even 1

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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 ) = 1 2 π + Z -∞ sin( t/ 2) t/ 2 e - itx dt We can set x = 0, and we get p (0) = 1 2 π + Z -∞ sin( t/ 2) t/ 2 dt = 1 π + Z -∞ sin z z dz replace z =
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