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

Assume the model $$\ y = X\beta + u \$$ with $$\ W \$$ is a $$\ n\times 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. $$\text{plim}_{n\to\infty} \ \left(\frac{X^TX}{n}\right) = m_{X^TX}$$

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

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

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

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

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

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

where we assume that $$\text{lim}_{n\to\infty} \left(\frac{{\mathbb E}\left(W^Tu\right)}{n}\right)$$ is equal to $$0$$.

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

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

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

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

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

$$\text{plim}_{n\to\infty} \ \left(\frac{X^TM_WX}{n}\right) = \left(\text{plim}_{n\to\infty} \ \frac{X^TX}{n}\right) - \left(\text{plim}_{n\to\infty} \ \frac{X^TP_WX}{n}\right)$$

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 $$\beta$$.

• I think that this analysis is correct. You could lose some efficiency and/or introduce bias in small samples, where you are doing extra work, but asymptotically it should not matter much if the information aligns the way you describe. – StasK Dec 10 '12 at 15:07
• The finite sample properties would be discussed after this small section. I just wanted to make sure that my description of the only case in which the positive semidefinite matrix is actually zero will be correct. – Druss2k Dec 10 '12 at 15:52