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EE562a_HW_2_Solutions

# EE562a_HW_2_Solutions - EE562A Homework 2 Solutions...

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EE562A Homework 2 Solutions Prepared by Hooman Shirani-Mehr Problem 1. Let X = ( X 1 , X 2 , X 3 , X 4 ) be a Gaussian random vector where E [ X i ] = 0 for i = 1 , 2 , 3 , 4 . Show that E [ X 1 X 2 X 3 X 4 ] = K 12 K 34 + K 13 K 24 + K 14 K 23 where K ij is the i, j element of the covariance matrix K X . Solution. Let x ( u ) be an n-dimensional vector of mean-zero real Gaussian random variables. The expected value of the product of the random variables in this vector can be computed by appropriate differentiation of the characteristic function of this Gaussian random vector. E h n Y t =1 x ( u, t ) i = ( - i ) n n ∂v 1 ...∂v n E h exp ± i n X t =1 v t x ( u, t ) ²i³ ³ ³ v = 0 The characteristic function can be substituted into this expression and the partial derivative calculated in an organized fashion. E h n Y t =1 x ( u, t ) i = ( - i ) n h n ∂v 1 ...∂v n exp( - v t K X v / 2) ³ ³ v = 0 = ( - i ) n h n ∂v 1 ...∂v n X m =0 ( - 1) m 2 m m ! ( - v t K X v ) m ³ ³ v = 0 = ( - i ) n X m =0 ( - 1) m 2 m m ! h n ∂v 1 ...∂v n ± n X j 1 =1 n X k 1 =1 v j 1 v k 1 K x ( j 1 , k 1 ) ² ... ± n X j m =1 n X k m =1 v j m v k m K x ( j m , k m ) ²i³ ³ ³ v = 0 (1) Now the differentiation process reduces to differentiating products of the form v j 1 v k 1 v j 2 v k 2 ...v j m v k m = n Y t =1 v n t t where n t simply counts the number of times that

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EE562a_HW_2_Solutions - EE562A Homework 2 Solutions...

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