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Is there an easy way to find the covariance between $x_t$ and $\epsilon_t^1$ in a system like

$$y_{t} = \beta x_{t} + \epsilon^1_{t}$$ $$x_{t} = \alpha y_{t-1} + \epsilon^2_{t},$$

potentially under the assumption that $\epsilon_t^1$ and $\epsilon_t^2$ have 0 covariance and each variance 1?

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  • $\begingroup$ How do you define $\beta$? Is $\beta:=\mathbb{E}(y_t|x_t)$ or $\beta:=\mathbb{E}(y_t|\text{do} \ x_t)$ or ...? I guess the answer may depend on that. $\endgroup$ – Richard Hardy Oct 30 '19 at 15:19
  • $\begingroup$ I do not understand $\beta:=\mathbb E[y_t \lvert x_t]$ did you mean $x_t\beta:=\mathbb E[y_t \lvert x_t]$ or perhaps $\beta:=\partial \mathbb E[y_t \lvert x_t]/\partial x_t$. I am not super familiar with Judea Pearl style do-operator. $\endgroup$ – Stop Closing Questions Fast Oct 31 '19 at 11:16
  • $\begingroup$ What I wrote was simply wrong. I guess I meant $\beta \ \text{such that} \ x_t\beta=\mathbb{E}(y_t|x_t)$. In that case $\epsilon^1\bot x$ by definition and their covariance is 0, is it not? I am also not very familiar with $\text{do}$, so even that part may be wrong. Should I perhaps have written $\beta \ \text{such that} \ x_t\beta=\mathbb{E}(y_t|\text{do}\ x_t)$? But the question is, what $\beta$ are you interested in? $\endgroup$ – Richard Hardy Oct 31 '19 at 11:47
  • $\begingroup$ Yes it follows from $\mathbb E[y_t \lvert x_t] = x_t \beta$ that $\mathbb cov[x_t\epsilon_t]=0$. I think I need to think more and then maybe update the question. $\endgroup$ – Stop Closing Questions Fast Oct 31 '19 at 13:08
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If you assume zero covariance between $\epsilon^1_t$ and $\epsilon^2_t$ as well as between $y_{t-1}$ and $\epsilon^2_t$, then $\beta$ is identified by a regression of $y_t$ on $x_t$ and $y_{t-1}$. This then identifies $cov(x_t, \epsilon^1_t) = cov(x_t, y_t - \beta x_t) = cov(x_t, y_t) - \beta var(x_t)$.

If you assume zero covariance between $\epsilon^1_t$ and $\epsilon^2_t$ as well as between $y_{t-1}$ and $\epsilon^1_t$, then $cov(x_t, e^1_t) = cov(\alpha y_{t-1} + \epsilon^2_{t}, \epsilon^1_t) = 0$.

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