# Implementation of 2SLS in regression with AR errors?

Consider the following simple example:

$$Y_t=\beta X_{t-1}+\varepsilon_t$$

$$X_t=\gamma Y_t + Z_t +u$$,

where

$$\varepsilon_t=\alpha\varepsilon_{t-1}+\eta$$,

and $$E[Z_t\varepsilon_t']=E[uu']=E[\eta\eta']=0$$.

Since $$X_{t-1}=\gamma Y_{t-1} + u=\gamma (\beta X_{t-2}+\alpha\varepsilon_{t-2}+\eta)+u$$

and

$$\varepsilon _t=\alpha\varepsilon_{t-1}+\eta_t=\alpha(\alpha \varepsilon_{t-2}+\eta_{t-1})+\eta_t$$

$$X_{t-1}$$ and $$\varepsilon_t$$ are clearly correlated through sharing $$\varepsilon_{t-2}$$.

Since regression with AR errors can simply apply OLS to obtain an unbiased slope parameter estimate (its s.d. notwithstanding), I'm wondering if 2SLS can be applied directly to obtain the same. To provide some context, I'm only interested in the magnitude of $$\beta$$, and GMM is computationally infeasible given the sample size.

• how can expectation of $E[uu']$ and $E[\eta \eta']$ be 0? – Jesper for President Oct 15 '19 at 20:37
• Also the argument $X_{t-1} = \gamma Y_{t-1} + u = \gamma (\gamma X_{t-2} + \alpha \epsilon_{t-2} + \eta)$ is as I read it flawed, (1) no time index on $\eta$ (2) should $\gamma$ on $X_{t-2}$ not be $\beta$ and $\epsilon_{t-2}$ be $\epsilon_{t-1}$ ... – Jesper for President Oct 15 '19 at 20:46
• You're right about the $\beta$, thanks, I'll correct. The $\varepsilon_{t-2}$ is correct. For the other concerns, they're just a short-hand since u and eta are not time dependent, and in full form I could write $E[u_i u_j']=E[u_t u_\tau']=0$. – seamux Oct 15 '19 at 23:09