# Covariance in system with lagged reverse causality

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?

• 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. – Richard Hardy Oct 30 '19 at 15:19
• 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. – Stop Closing Questions Fast Oct 31 '19 at 11:16
• 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? – Richard Hardy Oct 31 '19 at 11:47
• 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. – Stop Closing Questions Fast Oct 31 '19 at 13:08

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 $$cov(x_t, e^1_t) = cov(\alpha y_{t-1} + \epsilon^2_{t}, \epsilon^1_t) = 0$$.