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Define a symmetric Toeplitz matrix by $$\begin{pmatrix}c_1 & c_2 & c_3 & \cdots & c_n\\c_2 & c_1 & c_2 & \cdots & c_{n-1}\\c_3 & c_2 & c_1 & \cdots &c_{n-2}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\c_n & c_{n-1} & c_{n-2} & \cdots & c_1\end{pmatrix},$$ for $$c_1 = 1, \quad c_{k+1}=\frac{1}{2}\left((k+1)^{2-\alpha}-2 k^{2-\alpha}+(k-1)^{2-\alpha}\right) \quad (k>0),$$ where $\alpha \in (0, 1)$.

How to prove that this matrix is positive-definite? Thanks a lot.

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Let $B := \{ B_{t} \}_{t\geq 0} $ be a fractional Brownian motion of Hurst parameter $1-\alpha/2 $ on some probability space $(\Omega,\mathcal{F},P)$. Since $B$ has the covariance function $$ E(B_{t}B_{s}) = \frac{1}{2}\left( s^{2(1-\alpha/2)} + t^{2(1-\alpha/2)} - |t-s|^{2(1-\alpha/2)} \right), $$ a simple calculation yields $$ E( (B_{n+k}-B_{n+k-1})(B_{n}-B_{n-1}) ) = \frac{1}{2}\left( (k+1)^{2-\alpha} + |k-1|^{2-\alpha} - 2k^{2-\alpha} \right) $$ for any integers $n\geq 1$ and $ k\geq 0$ ($E$ means the expectation under $P$). Let us set (note that $B_{0}=0$ a.s.) $$ X=\left[ \begin{array}{c} B_{1} \\ B_{2} \\ \cdots \\ B_{k} \end{array}\right],\ \Delta X=\left[ \begin{array}{c} B_{1}-B_{0} \\ B_{2}-B_{1} \\ \cdots \\ B_{k}-B_{k-1} \end{array}\right],\ A=\left[ \begin{array}{c} 1 & 0 & 0 & \cdots & 0 \\ -1 & 1 & 0 & \cdots & 0 \\ 0 & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots &\ddots & \ddots & 0 \\ 0 & \cdots &0 & -1 & 1 \end{array}\right], $$ then the symmetric Toeplitz matrix of interest (which would be denoted by $C$) can be written as $$ C = E\left( \Delta X(\Delta X)^{\top} \right)= A E(XX^{\top})A^{\top}, $$ where the superscript $\top$ means transpose. Since $E(XX^{\top})$ is positive-definite (a proof of this fact can be found in, e.g., proposition 1.6 of "Selected Aspects of Fractional Brownian Motion" by Ivan Nourdin), $C$ is also positive-definite.

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  • $\begingroup$ @Cave Johnson: Thank you for your pointing out. But I think I surely provided a proof that the symmetric Toeplitz matrix in the question (which is denoted by C in my answer) is positive-definite. $\endgroup$ – Yurichlet Jan 9 '17 at 16:04
  • $\begingroup$ I see :) I hadn't read it through, but only given a first sight the time I post the comment. I'm sorry for that. $\endgroup$ – Cave Johnson Jan 10 '17 at 4:50

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