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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 $\text{Cov}(X_t, X_{t-j})$?

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  • $\begingroup$ I am asking what is the general formula for covariance, not autocovariance $\endgroup$ Commented Mar 5, 2017 at 0:46
  • $\begingroup$ What does that mean in this context? $\endgroup$ Commented Mar 5, 2017 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. $\endgroup$ Commented Mar 5, 2017 at 1:03
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    $\begingroup$ 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. $\endgroup$ Commented Mar 5, 2017 at 10:32
  • $\begingroup$ Related: stats.stackexchange.com/questions/325137/… $\endgroup$ Commented Jun 17, 2020 at 14:27

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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: $$C_o = C_i\, \star\, \left(\phi(x)\, \star\, \phi(x^{-1})\right).$$

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    $\begingroup$ 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. $\endgroup$ Commented Oct 21, 2017 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
    Commented Apr 11, 2019 at 4:56

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