Assume we have a regression model \begin{equation} y_{i} = z_{i}' \delta + \epsilon_{i} \end{equation} 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!

  • 1
    $\begingroup$ Add the self study tag. $\endgroup$ – Michael Chernick Jan 22 at 21:20

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