… Another potential problem with applying 2SLS and other IV procedures
is that the 2SLS standard errors have a tendency to be ‘‘large.’’ What
is typically meant by this statement is either that 2SLS coefficients
are statistically insignificant or that the 2SLS standard errors are
much larger than the OLS standard errors. Not surprisingly, the
magnitudes of the 2SLS standard errors depend, among other things, on
the quality of the instrument(s) used in estimation.
This quote is from Wooldridge's "Econometric analysis of cross-sectional and panel data". I wonder why this happens? I would prefer a mathematical explanation.
Assuming homoskedasticity for simplicity the (estimated) asymptotic variance of OLS estimator is given by
$$\widehat{Avar}(\hat{\beta}_{OLS}) = n\sigma^2(X'X)^{-1}$$
while for the 2SLS estimator
$$\widehat{Avar}(\hat{\beta}_{2SLS}) = n\sigma^2(\hat{X}'\hat{X})^{-1}$$
where
$$\hat{X} = P_zX = Z(Z'Z)^{-1}Z'X.$$
$X$ is the matrix of regressors, including the endogenous ones, and $Z$ is the matrix of instrumental variables.
So rewriting the variance for 2SLS gives
$$\widehat{Avar}(\hat{\beta}_{2SLS}) = n\sigma^2\left(X'Z(Z'Z)^{-1}Z'X\right)^{-1}.$$
However, I can't conclude from above formulas that $\widehat{Avar}(\hat{\beta}_{2SLS}) \geq \widehat{Avar}(\hat{\beta}_{OLS})$.
Best Answer
We say a matrix $A$ is at least as large as $B$ if their difference $A-B$ is positive semidefinite (psd).
An equivalent statement that turns out to be handier to check here is that $B^{-1}-A^{-1}$ is psd (much like $a>b$ is equivalent to $1/b>1/a$).
So we want to check that $$ X'X-X'Z(Z'Z)^{-1}Z'X $$ is psd.
Write $$ X'X-X'Z(Z'Z)^{-1}Z'X=X'(I-Z(Z'Z)^{-1}Z')X=X'M_ZX $$ To check that $X'M_ZX$ is psd, we must show that, for any vector $d$, $$ d'X'M_ZXd\geq0 $$ Let $c=Xd$. Then, $$ c'M_Zc\geq0 $$ as $M_Z$ is a symmetric and idempotent projection matrix, which is known to be psd: write, using symmetry and idempotency, $$ c'M_Zc=c'M_ZM_Zc=c'M_Z'M_Zc $$ and let $e=M_Zc$, so that $c'M_Zc=e'e=\sum_ie_i^2$, which, being a sum of squares, must be nonnegative.
P.S.: Two little quibbles - you refer to the estimated asymptotic variances $\widehat{Avar}(\hat\beta_j)$. Now, the OLS estimator and the 2SLS estimator of $\sigma^2$ are not the same, so that I do not see that the ranking must necessarily be preserved if these estimates differ. Also, the asymptotic variances are generally scaled by $n$ so as to obtain a nondegenerate quantity as $n\to\infty$. (Of course, scaling both by $n$ will not affect the ranking, so that the issue is a little moot for this particular question.)