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  • $\begingroup$ Thank you for the answer. In this case, the answer to this question is wrong: stats.stackexchange.com/questions/196994/… To show the variance of $r_t=\alpha + \phi_1 r_{t-1} + \theta_1 a_{t-1} + a_{t}$, he makes use of the fact that $E(a_t r_{t-1})=0$. $\endgroup$ Commented Dec 29, 2021 at 20:09
  • $\begingroup$ @Fam, that answer is not wrong, since it is $r$ that is lagged w.r.t. $a$ and not the other way around. You can observe this visually from the CCF graph I have included, or you could show it algebraically. $\endgroup$ Commented Dec 29, 2021 at 20:36
  • $\begingroup$ So, $\varepsilon_t$ and $X_{t-1}$ are not correlated, in my question? Suppose that in the question in my first comment, we have $\alpha =0$. So, $E(r_t)=0$ and $E(r_{t-1})=0$. Thus, $cov(a_t,r_{t-1} ) = E(a_t r_{t-1})=0$. $\endgroup$ Commented Dec 29, 2021 at 20:48
  • $\begingroup$ I dont underestand why $X_t$ and $\varepsilon_{t-1}$ are positive correlated, but $X_{t-1}$ and $\varepsilon_t$ are not? $\endgroup$ Commented Dec 29, 2021 at 20:50
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    $\begingroup$ @Fam, your intuition is correct. $\varepsilon_t$ is generated independently of past values of $X$, but $X_t$ is a linear function of past values of $\varepsilon$. $\endgroup$ Commented Dec 29, 2021 at 21:12