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.


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 $0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.