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By Falk M.

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16)   1 −k (−k)2 . . (−k)p 1 −k + 1 (−k + 1)2 . . (−k + 1)p    X = .  ..   .. . 2 p 1 k k ... 17) is the design matrix, β = (β0 , . . , βp )T and y = (yt−k , . . , yt+k )T . The rank of X T X equals that of X, since their null spaces coincide (Exercise 11). Thus, the matrix X T X is invertible iff the columns of X are linearly independent. But this is an immediate consequence of the fact that a polynomial of degree p has at most p different roots (Exercise 12). 16) have, therefore, the unique solution β = (X T X)−1 X T y.

Zp ∈ C of A(z) = 1 + a1 z + a2 z 2 + . . , |zi | > 1 for 1 ≤ i ≤ p. Proof. We know from the Fundamental Theorem of Algebra that the equation A(z) = 0 has exactly p roots z1 , . . g. 5 in Conway (1975)), which are all different from zero, since A(0) = 1. g. 6 in Conway (1975)) A(z) = ap (z − z1 ) · · · (z − zp ) z z z 1− ··· 1 − , = c 1− z1 z2 zp where c := ap (−1)p z1 · · · zp . 1 Linear Filters and Stochastic Processes 51 where the coefficients (1/zi )u , u ≥ 0, are absolutely summable.

Ap and au = 0 elsewhere has an absolutely summable and causal inverse filter if the p roots z1 , . . , zp ∈ C of A(z) = 1 + a1 z + a2 z 2 + . . , |zi | > 1 for 1 ≤ i ≤ p. Proof. We know from the Fundamental Theorem of Algebra that the equation A(z) = 0 has exactly p roots z1 , . . g. 5 in Conway (1975)), which are all different from zero, since A(0) = 1. g. 6 in Conway (1975)) A(z) = ap (z − z1 ) · · · (z − zp ) z z z 1− ··· 1 − , = c 1− z1 z2 zp where c := ap (−1)p z1 · · · zp . 1 Linear Filters and Stochastic Processes 51 where the coefficients (1/zi )u , u ≥ 0, are absolutely summable.

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