# Bootstrap confidence interval for just-identified IV estimator

Assume we have a regression model $$$$y_{i} = z_{i}' \delta + \epsilon_{i}$$$$ with dependent variable $$y_{i}$$, L regressors $$z_{i}$$ and K instruments $$x_{i}$$, and assumptions of 1) linearity, 2) ergodic stationarity, 3) pre-determined instruments ($$E(x_{i} \epsilon_{i}$$)=0), 4) $$E(x_{i} z_{i})$$ has full column rank and 5) $$x_{i} \epsilon_{i}$$ is a martingale difference sequence.

Then the GMM estimator estimates $$\delta$$ by the solution of the sample analogue to the moment condition $$\bar{g} = 0$$. This can be re-written such that $$\bar{g}=X'Y-X'Z \delta$$.

Now, in the just-identified case, when K=L, this can be solved for delta, such that we get the IV estimator $$\hat{\delta_{IV}}=(X'Z)^{-1}X'y$$.

A question on an old exam of a course I am studying asked: 'describe how you would obtain a bootstrap 95% confidence interval for this estimator'. My confusion is: since the system of equation is just-identified, we can obtain an exact solution. Then why do we need a confidence interval? My guess is because the exact solution is still only a solution to a sample moment condition? So then would it be correct to say we construct the bootstrap CI by re-drawing multiple bootstrap samples from the original sample and re-estimating the IV estimator, then taking the symmetric bounds of all the bootstrap estimates around the original estimate that comprise 95% of the bootstrap estimates? Thank you!

• Add the self study tag. – Michael Chernick Jan 22 at 21:20