By D'Agostini G.

Those notes are in response to seminars and minicourses given in quite a few areas during the last 4 years.

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43) 4. 41) may not exist. For any ﬁnite θ, however, the following similar fact is true: If F (x) is a continuous function of [1/Mf , θ], then lim n−1 n→∞ n−1 F [min(ρn,k , θ)] = (2π)−1 k=0 2π F [min(1/f (λ), θ)] dλ. 44) Proof. 1. 9) 2 1 π n−1 ∗ ikλ x Tn x = xk e f (λ)dλ > 0 2π −π k=0 so that for all n min τn,k > 0 k and hence n−1 det Tn = τn,k = 0 k=0 so that Tn (f ) is nonsingular. 2. 1 since f (λ) ≥ mf > 0 ensures that Tn−1 , Cn−1 ≤ 1/mf < ∞. 40 CHAPTER 4. 45) must hold for large enough n. 40), however, if n is large enough, then probably the ﬁrst term of the series is suﬃcient.

The proof of part 4 is motivated by one of Widom [2]. Further results along the lines of part 4 regarding unbounded Toeplitz matrices may be found in [5]. , ﬁnding conditions on f (λ) to ensure that Tn (f ) is invertible. 42 CHAPTER 4. TOEPLITZ MATRICES Parts (a)-(d) can be straightforwardly extended if f (λ) is continuous. For a more general discussion of inverses the interested reader is referred to Widom [2] and the references listed in that paper. It should be pointed out that when discussing inverses Widom is concerned with the asymptotic behavior of ﬁnite matrices.

2 Let Xn be an autoregressive process with covariance matrix (n) RX with eigenvalues ρn,k . Then (RX )−1 ∼ σ −2 Tn (|a|2 ). 24) where 1/ρn,k are the eigenvalues of (RX )−1 . 25) so that the process is asymptotically stationary. Proof. ) Note that if |a(λ)|2 > 0, then 1/|a(λ)|2 is the spectral density of Xn . If (n) |a(λ)|2 has a zero, then RX may not be even asymptotically Toeplitz and hence Xn may not be asymptotically stationary (since 1/|a(λ)|2 may not be integrable) so that strictly speaking xk will not have a spectral density.