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I wonder if there is a paper that can point out how to compute the determinant of a $d \times d$ autoregressive correlation matrix of the form

$$R = \begin{pmatrix} 1 & r & \cdots & r^{d-1}\\ r & 1 & \cdots & r^{d-2}\\ \vdots & \vdots & \ddots & \vdots\\ r^{d-1} & r^{d-2} & \cdots & 1 \end{pmatrix}$$

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The determinant of this matrix is $(1-r^2)^{d-1}$. This can be proven by induction on $d$ using the block matrix identity $\det \pmatrix{A & B\cr C & D} = \det(A) \det(D - C A^{-1} B)$, where $A$ is the top left $(d-1) \times (d-1)$ submatrix. Note that $$A \pmatrix{0\cr \ldots\cr 0\cr r\cr} = B$$ so $D - C A^{-1} B = 1 - r^2$.

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  • $\begingroup$ This is a correct answer I misread the OP $\endgroup$ – Henry.L Jul 19 '17 at 21:57
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The method of Dodgson's evaluation of determinants offers a simple inductive (recursive) proof.

If we denote your determinant by $R_d$, the above technique leads to $$R_d=\frac{R_{d-1}^2-0^2}{R_{d-2}}=\frac{(1-r^2)^{2d-4}}{(1-r^2)^{d-3}}=(1-r^2)^{d-1},$$ since the top right-most (and bottom left-most) matrices of size $(d-1)\times(d-1)$ have vanishing determinant.

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A thorough discussion is contained in

Finch, P. D. "On the covariance determinants of moving-average and autoregressive models." Biometrika 47.1/2 (1960): 194-196. JSTOR

But it is more common to write it in form of product of conditional densities $f(x_2\mid x_1)f(x_3\mid x_1,x_2)\cdots $ and consider each update sequentially to discuss the behavior of the covariance matrix.

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