Solved – When are the asymptotic variance of OLS and 2SLS equal

Assume the model $ y = Xbeta + u $ with $ W $ is a $ ntimes l $ so called matrix of instruments.

The following assumptions hold. There is a law of large numbers (LLN) for 1.,2.,3. and 4. such that

  1. $text{plim}_{ntoinfty} left(frac{X^TX}{n}right) = m_{X^TX}$

    holds where $m_{X^TX}$, a finite, non stochastic matrix with full
    column rank, exsists.

  2. $text{plim}_{ntoinfty} left(frac{W^TW}{n}right) =
    text{lim}_{ntoinfty} left(frac{{mathbb

    where we assume, that the RHS exists and is finite as well as
    positive definite.

  3. $text{plim}_{ntoinfty} left(frac{W^TX}{n}right)$

    where we assume, that the limit $W^TX$ exists, is finite and $W^TX$
    has full column rank i.e. $text{rk}left(W^TXright)=k$.

  4. $text{plim}_{ntoinfty} left(frac{W^Tu}{n}right) =
    text{lim}_{ntoinfty} left(frac{{mathbb
    E}left(W^Turight)}{n}right) = 0$

    where we assume that $text{lim}_{ntoinfty} left(frac{{mathbb
    is equal to $0$.

The asymptotic variance for OLS will, under the assumption of homoscedastic errors $u$ i.e. that ${mathbb E}left(uu^T|Xright) = sigma^2_0 I_n$ and that the model is actually correctly specified , be

$text{plim}_{ntoinfty} text{Var}left[ n^{frac{1}{2}}left(boldsymbol{widehat{beta}}_{text{KQ}} – boldsymbol{beta_0}right)big| Xright] = sigma_0^2 m_{X^TX}^{-1}$

The corresponding asymptotic variance for the 2SLS case will look like

$text{plim}_{ntoinfty} text{Var}left[ n^{frac{1}{2}}left(boldsymbol{widetilde{beta}_{text{2SLS}}} – boldsymbol{beta_0}right)big| Xright] = sigma_0^2 text{plim}_{ntoinfty} left(frac{X^TP_W X}{n}right)^{-1}$

If I now consider the precision matrix rather then the variance matrix and take their difference I'll get

$text{plim}_{ntoinfty} left(frac{X^TM_WX}{n}right) = left(text{plim}_{ntoinfty} frac{X^TX}{n}right) – left(text{plim}_{ntoinfty} frac{X^TP_WX}{n}right)$

which is a positive semidefinite matrix.

My question: When are both asymptotic variances of OLS and 2SLS equal

My guess is that we need to set $X^TX = X^T P_W X$. The only possibility for such a thing is when $P_WX=X$. So all the columns of $X$ must be in the image of $P_W$. This means that all the columns of $X$ are actually viable instruments. By that I conclude that there are (at least asymptotically) no endogenous regressor within $X$ if their asymptotic variances are equal. But this seems to good to be true. There are downfalls of using a 2SLS-approach if theres no endogeneity involved. Again, this would mean that if the variances of the 2SLS and OLS estimator are equal asymptotically it does not matter if we choose 2SLS or OLS since both estimators are consistent for $beta$.

To close this one:

When are the asymptotic variances of OLS and 2SLS equal?

A: Only when the "matrix of instruments" essentially contains exactly the original regressors, (or when the instruments predict perfectly the original regressors, which amounts to the same thing), as the OP himself concluded.

The variance of the IV estimator (generalized or not), when the OLS is also consistent, is never smaller then the variance of the OLS.

This is one of the main reasons why the "Hausman specification test" (or the Durbin-Wu-Hausman) is performed: because if the data show that the two estimators have the same probability limits (i.e if the null of the test is not rejected), then we should stick to OLS, which has a lower variance.

If the null is rejected (and we have other arguments that support the consistency of the IV estimator), then the OLS is deemed inconsistent, and then we go with the IV estimator, to buy consistency at the price of higher variance.

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