Is the hilbert space generated by the linear span of

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is the Hilbert space generated by the linear span of Y t .) Let’s now solve min ˆ X t ∈H y E [( X t - ˆ X t ) 2 ] . ((*)) A couple important properties: If E [ X 2 t ] < then ˆ X t ∈ H y solves (*) if and only if E [( X t - ˆ X t ) Z ] = 0 for all Z ∈ H y . That is, the error is orthogonal to all elements of H y . Proof “If”: Suppose ˆ X t ∈ H y satisfies E [( X t - ˆ X t ) Z ] = 0 for all Z ∈ H y . Let X * t be an element of H y . E [( X t - X * t ) 2 ] = E [( X t - ˆ X t + ˆ X t - X * t ) 2 ] = E [( X t - ˆ X t ) 2 ] + 2 E [( X t - ˆ X t ) ( ˆ X t - X * t ) | {z } ∈H t ] + E [( ˆ X t - X * t ) 2 ] = E [( X t - ˆ X t ) 2 ] + E [( ˆ X t - X * t ) 2 ] E [( X t - ˆ X t ) 2 ] . So the orthogonality condition is sufficient for achieving MMSE. “Only if”: Suppose ˆ X t ∈ H y , and there is an element Z ∈ H y such that E [( X t - ˆ X t ) Z ] 6 = 0 . We will show that there would then be a better estimate: Let X * t = ˆ X t + E [( X t - ˆ X t ) Z ] E [ Z 2 ] . Then E [( X t - X t * ) 2 ] = E [( X t - ˆ X t ) 2 ] - ( E [( X t - ˆ X t ) Z ]) 2 E [ Z 2 ] < E [( X t - X * t ) 2 ] . So ˆ X t cannot be the MMSE estimator, which implies the necessity of the orthogo- nality condition. 2 E [( X t - ˆ X t ) Z ] = 0 for all Z ∈ H y if and only if E [ ˆ X t ] = E [ X t ] and E [( X t - ˆ X t ) Y τ ] = 0 for all τ [ a, b ] . This is a restatement of orthogonality, but for a restricted space. Proof “Only if” (necessity): Want to show that E [( X t - ˆ X t ) Z ] = 0 only if E [( X t - ˆ X t )] = 0 and E [( X t - ˆ X t ) Y τ ] = 0 for all τ [ a, b ] . But this comes by definition, since 1 ∈ H y and Y τ ∈ H y for each τ [ a, b ] . “If”: (sufficiency) Suppose Z ∈ H y and E [ X t - ˆ X t ] = 0 and E [( X t - ˆ X t ) Y τ ] = 0 for all τ [ a, b ] . (That is, the error is orthogonal to each Y τ .) Then for Z = lim n →∞ ( MS )( n X i =1 a i Y t i + c ) we have E [( X t - ˆ X t ) Z ] = lim n →∞ E [( X t - ˆ X t )( n X i =1 a i Y t i + c )] ,
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ECE 6010: Lecture 9 – Linear Minimum Mean-Square Error Filtering 3 where the limit may be interchanged because X t is assumed to be second order, E [( X t - ˆ X t ) Z ] = lim n →∞ n X i =1 a i E [( X t - ˆ X t ) Y t i ] + cE [( X t - ˆ X t )] = 0 . 2 Suppose we further restrict ˆ X t to be of the form ˆ X t = Z b a h ( t, τ ) Y τ + c t . That is, ˆ X t is the output of a linear filter driven by Y t . Note that ˆ X t ∈ H y . By property 2, we must have (1) E [ X t ] = E [ ˆ X t ] = Z b a h ( t, τ ) μ Y ( τ ) + c t so that c t = μ x ( t ) - Z b a h ( t, τ ) μ y ( τ ) and (2): E [ X t Y τ ] = E [ ˆ X t Y τ ] for τ [ a, b ] . That is, R XY ( t, τ ) = Z b a h ( t, σ ) R Y ( σ, τ ) + c t μ y ( τ ) This gives us two equations in the unknowns c t and h . We can eliminate c t : R XY ( t, τ ) = Z b a h ( t, σ )( R Y ( σ, τ
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  • Fall '08
  • Stites,M
  • Xt, Wiener filter, Fredholm integral equation

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