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Let $\{X_t\}_{t\in\mathbb{Z}}$ is the stacionary autoregressive process of degree p (AR(p)), and autocorrelation function of AR(p) is $$\rho(s)=\phi_1\rho(s-1)+\phi_2\rho(s-2)+\dots+\phi_p\rho(s-p), \text{ for $s=1,2,\dots, p$}.$$ I should show that when $s\rightarrow \infty$ then $\rho(s)\rightarrow0$. I am trying to solve difference equation and then $s\rightarrow \infty$. But I have no idea how to make it in general, because I think that when $s$ goes to the infinity ($s\rightarrow \infty$), it means also that p goes to the infinity ($p\rightarrow \infty$). Any help will be appreciated, thank you very much.

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This can be seen by solving for $\{\rho(s)\}_{s \geq 0}$ directly.

The system of linear homogeneous difference equation (of degree $p$) is $$ \underbrace{ \rho(s)- \phi_1\rho(s-1)+\phi_2\rho(s-2)+\dots+\phi_p\rho(s-p) }_{ \Phi(L)\rho(s)} = 0, \;\; s \geq {0}. $$ with initial conditions $\rho(0), \cdots \rho(p-1)$ given by the Yule-Walker equations

$$ \begin{bmatrix} 1 &\rho(1) & \rho(2) &\cdots& \rho(p-1) \\ \rho(1) & 1 & \rho(1) &\cdots & \rho(p-2) \\ \rho(2) & \rho(1) & 1 &\cdots & \rho(p-3) \\ \vdots & \vdots & \vdots &\ddots & \vdots \\ \rho(p-1) & \rho(p-2) &\rho(p-3)&\cdots & 1 \\ \end{bmatrix} \begin{bmatrix} \phi_1 \\ \phi_2 \\ \phi_3 \\ \vdots \\ \phi_p \\ \end{bmatrix} = \begin{bmatrix} \rho(1) \\ \rho(2) \\ \rho(3) \\ \vdots \\ \rho(p) \\ \end{bmatrix}. $$

The general solution is a linear combination of terms corresponding to the AR polynomial $\Phi(z)$.

For example, if $\Phi$ have $p$ distinct real roots $r_1, \cdots, r_{p}$, the general solution is $$ \rho(s) = c_1 r_1^{-s} + \cdots c_{p} r_{p}^{-s}. $$ The coefficients $c_1, \cdots, c_{p}$ are given by the initial conditions. By assumption (covariance stationarity), $|r_1|, \cdots, |r_{p+1}| > 1$, $$ \lim_{s \rightarrow \infty} r_l^{-s} = 0, \; l = 1, \cdots, p. $$ This implies $$ \lim_{s \rightarrow \infty} \rho(s) = 0. $$

The case of repeated or complex roots is similar. Same result holds.

The key condition is that all roots of $\Phi$ lies outside the unit circle. If there is a root on or inside the unit circle, e.g. if $r =1$ is a root, then the term $1^{-s}$ in $\rho(s)$ clearly does not approach $0$.

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