1
$\begingroup$

I am not sure what the formula is for the covariance of an AR(2) process $$ X_t = \phi_1 X_{t−1} + \phi_2 X_{t−2} + \epsilon_t $$ where $\{\epsilon_t\}$ is white noise process (Gaussian) $N(0, \sigma^2)$.

What is the formula for $\text{Cov}(X_t, X_{t-j})$?

$\endgroup$
  • $\begingroup$ I am asking what is the general formula for covariance, not autocovariance $\endgroup$ – user7543128 Mar 5 '17 at 0:46
  • $\begingroup$ What does that mean in this context? $\endgroup$ – Michael R. Chernick Mar 5 '17 at 0:47
  • $\begingroup$ Okay but note that it exists only if the roots of the characteristic polynomial fall outside the unit circle in the complex plane. $\endgroup$ – Michael R. Chernick Mar 5 '17 at 1:03
  • 1
    $\begingroup$ Check out the existing posts on the subject at Cross Validated. You will find either an exact same question or similar ones that will help you. $\endgroup$ – Richard Hardy Mar 5 '17 at 10:32
  • $\begingroup$ Related: stats.stackexchange.com/questions/325137/… $\endgroup$ – Christoph Hanck Jun 17 at 14:27
1
$\begingroup$

Just use the bilinearity of the covariance function: $$\operatorname{cov}\left(\sum_{i=1}^n a_iX_i, \sum_{j=1}^m b_j Y_j\right) = \sum_{i=1}^n\sum_{j=1}^m a_i b_j(\operatorname{cov}X_i, Y_j)$$ to get \begin{align} \operatorname{cov}(X_t,X_{t-j}) &= \operatorname{cov}(\phi_1X_{t-1}+\phi_2X_{t-2}+\epsilon_t, \phi_1X_{t-j-1}+\phi_2X_{t-j-2}+\epsilon_{t-j})\\ &= \phi_1^2 \operatorname{cov}(X_{t-1},X_{t-j-1}) + \phi_1^2 \operatorname{cov}(X_{t-2},X_{t-j-2})\\ & \quad +\phi_1\phi_2 \operatorname{cov}(X_{t-1},X_{t-j-2}) + \phi_1\phi_2 \operatorname{cov}(X_{t-2},X_{t-j-1}) \end{align} which gives a linear recurrence relationship that you can try and solve for yourself, or read up on how to go about doing so in your book. A very general relationship for time series defined by linear recurrences is that the "output" autocovariance function is given by the convolution of the "input" autocovariance function and the "autocovariance" of the characteristic function of the recurrence: $$C_o = C_i\, \star\, \left(\phi(x)\, \star\, \phi(x^{-1})\right).$$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ For $j>0$, the covariances of $\epsilon_{t-j}$ with $X_{t-1}$ and $X_{t-2}$ are non-zero so your second equation is incorrect. $\endgroup$ – Jarle Tufto Oct 21 '17 at 8:14
  • $\begingroup$ The linear recurrence relationship mentioned by @Dilip Sarwate should be \begin{align} \operatorname{cov}(X_t,X_{t-j}) &= \operatorname{cov}(\phi_1X_{t-1}+\phi_2X_{t-2}+\epsilon_{t}, X_{t-j}) \\ &= \phi_{1}\operatorname{cov}(X_{t-1}, X_{t-j}) + \phi_2 \operatorname{cov}(X_{t-2}, X_{t-j}) + \operatorname{cov}(\epsilon_{t}, X_{t-j}) \\ &= \phi_{1}\operatorname{cov}(X_{t-1}, X_{t-j}) + \phi_2 \operatorname{cov}(X_{t-2}, X_{t-j}) \end{align} $\endgroup$ – nalzok Apr 11 '19 at 4:56

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.