# Bias of Instrumental Variables Estimator

Given the IV estimator for $$\beta_1$$ (From $$y = \beta_0 +\beta_1x + u$$):

$$\beta_1 = \frac{Cov(z, y)}{Cov(z, x)}$$

And it's sample analog:

$$\hat{\beta_1} = \frac{\hat{Cov}(z, y)}{\hat{Cov}(z, x)}$$

Which simplifies to:

$$\hat{\beta_1} = \frac{\sum_{i=1}^n(z_i-\bar{z})(\beta_0 +\beta_1x_i + u)}{\sum_{i=1}^n(z_i - \bar{z})(x_i - \bar{x})} \rightarrow \beta_1 + \frac{\sum_{i=1}^n(z_i - \bar{z})u_i}{\sum_{i=1}^n(z_i - \bar{z})(x_i - \bar{x})}$$

Surely this estimator is unbiased because $$Cov(z, u)=0$$ by the assumption of exogeneity when dealing with instruments? Even if the above represents the sample covariance, because the sample covariance is an unbiased estimator of the population covariance, then surely $$\hat{\beta_1}$$ is still unbiased?

Aside from that, why is it necessary to take the conditional expectation on both $$X$$ and $$Z$$?.

$$E[\hat{\beta_1}] = E[E[\hat{\beta_1}|X, Z]] \rightarrow \frac{\sum_{i=1}^n(z_i - \bar{z})E[u_i|X, Z]}{\sum_{i=1}^n(z_i - \bar{z})(x_i - \bar{x})}$$

I can see why this expression isn't equal to zero (Because of the implication of the conditional on $$X$$), but why do we have to take that expectation instead of simply $$E[\hat{\beta_1}] = E[E[\hat{\beta_1}|X]]$$ or $$E[E[\hat{\beta_1}|Z]]$$? Is it necessary to condition on all of the data within the model?

The IV estimator you mention is a special case of two-stage least squares (TSLS) for a single endogenous variable ($$x$$). TSLS is asymptotically unbiased. It has a bias of order $$\mathcal{O}(\frac{1}{n})$$ (c.f. Fuller, 1977, Theorem 1 for $$\alpha=0$$). Fuller (1977) proposed an adjustment of the TSLS that has a bias of order $$\mathcal{O}(\frac{1}{n^2})$$ (c.f. Fuller, 1977, Theorem 1 for $$\alpha=1$$). Note that when people refer to the Fuller(1) estimator, they typically refer to an adjustment to the limited information maximum likelihood (LIML), which is the MLE of the causal parameter assuming Gaussian perturbations.