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$$y_i =\beta_0 +\beta_1 x_{1i}+\beta_2 x_{2i}+\varepsilon_i$$

Suppose $x_1$ is endogenous but $x_2$ is exogenous. We have two excluded instruments, $z_1$ and $z_2$. The first stage is:

$$x_{1i} =\delta_0 +\delta_1 z_{1i}+\delta_2 z_{2i}+\delta_3 x_{2i}+\eta_i $$

One method of estimation is 2SLS. Another option is LIML which exploits an assumption of joint normality of $\eta_i$ and $\varepsilon_i$ and is the maximum likelihood estimator.

It is documented in reputable sources that LIML and 2SLS have the same asymptotic distribution. For example, Bruce Hansen's textbook on page 349 says "The LIML estimator has the same asymptotic distribution as 2SLS". Green's textbook (7th edition), footnote 46 on page 372 confirms this.

Thus, the 2SLS estimator attains the same asymptotic variance matrix as the MLE estimator. With this, I want to say "2SLS is asymptotically equivalent to MLE".

Am I overlooking something? It doesn't feel right because MLE makes the stronger assumption of normality. It feels like with that additional assumption there should be an efficiency gain.

Michael Gmeiner
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    I remember learning in class that 2SLS can be reinterpreted as pseudo MLE because it can be showed that FOC of MLE estimator are the same as the ones defining 2SLS if you assume normal distribution for the error, but I can't find my notes with explanation anymore. However, that would make them same with that additional assumption but not necessarily without it. – 1muflon1 Mar 23 '22 at 17:40
  • Efficiency is about how well you employ the information in a given sample. It's not a useful concept to think about asymptotics where the sample size goes to infinity. MLE will be the most efficient regardless of the sample size if the shocks are normally distributed. 2SLS will not be equivalent unless the sample size goes to infinity. – jpfeifer Mar 24 '22 at 08:55

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