Equating the constants in front we must have σ 2 x

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Equating the constants in front we must have σ 2 x σ 2 y (1 - ρ 2 ) = s 11 s 22 - s 2 21 . It is clear that μ x and μ y are the first and second elements of μ . Let us assume only for convenience that μ = 0 . Expanding the exponent of the second form of the pdf and equating to terms in the first form of the pdf we find s 11 - s 2 21 /s 22 = σ 2 x (1 - ρ 2 ) s 22 - s 2 21 /s 11 = σ 2 y (1 - ρ 2 ) 4
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s 11 s 22 /s 12 - s 12 = σ x σ y Comparing all of these, we have σ 2 x = s 11 σ 2 y = s 22 ρ = s 21 s 11 s 22 5. Suppose X = h X 1 X 2 X 3 i ∼ N ( μ , Σ) where μ = 1 2 3 Σ = 4 2 1 2 6 3 1 3 8 (a) The value X 1 = 1 . 5 is measured. Determine the best estimate for ( X 2 , X 3 ). We can write μ 2 , 3 for the mean of the unmeasured variables. The estimation formula is ˆ μ 2 , 3 = μ 2 , 3 + Σ [2 , 3] , 1 Σ - 1 1 ( x 1 - μ 1 ) where Σ [2 , 3] , 1 is the covariance of the unmeasured values with the measured value, Σ [2 , 3] , 1 = cov( X 2 , X 1 ) cov( X 3 , X 1 ) = 2 1 and Σ 1 = Σ 1 , 1 = 4. We obtain ˆ μ 2 , = 2 3 + 2 1 (4) - 1 (1 . 5 - 1) = 2 . 25 3 . 125 . (b) In a separate problem, the values X 2 = 1 and X 3 = 5 are measured. Determine the best estimate of X 1 . ˆ μ 1 = μ 1 + Σ 1 , [2 , 3] Σ - 1 [2 , 3] ( x 2 x 3 - μ 2 , 3 ) where Σ 1 , [2 , 3] = cov( X 2 , X 1 ) cov( X 3 , X 1 ) = 2 1 and Σ [2 , 3] = 6 3 3 8 Thus ˆ μ 1 = 1 + 2 1 6 3 3 8 - 1 ( 1 5 - 2 3 ) = 0 . 6667 . 5
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(c) Determine a random vector Y which is a whitened version of X . Using Matlab , S = [4 2 1; 2 6 3; 1 3 8]; R = chol(S); C = R’; C*C’ % check the result we determine that Σ = 2 0 0 1 2 . 2361 0 0 . 5 1 . 1180 2 . 5495 2 0 0 1 2 . 2361 0 0 . 5 1 . 1180 2 . 5495 T = CC T .
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  • Fall '08
  • Stites,M
  • Randomness, σx σy, |Y, fX |Y

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