For OLS, my professor said that you should always test for heteroscedasticity first, rather than going straight to the adoption of robust standard errors.

I didn't quite follow this and no explanation was given.

I know heteroscedasticity doesn't alter the unbiasedness of OLS, but SE(Beta) is no longer consistent for the standard deviation.

Any help?


Standard errors affect your t-tests and F-tests and your confidence intervals. To treat the regression as violating one of the stochastic assumptions while it doesn't (i.e. use robust SE while heteroskedasticity is absent), is no less of a mistake than treating the regression as not violating the assumption while it does (i.e. use non-robust SE while heteroskedasticity is present).

Formally, for the regression specification

$$y_i = \mathbf x_i'\beta + u_i,\;\;\ i=1,...,n$$ the theoretical expression for the asymptotic variance of the OLS estimator (to be exact, of $\sqrt n(\hat \beta - \beta)$), for robust standard errors is

$$ \Big[E(\mathbf x_i \mathbf x_i')\Big]^{-1}E(u_i^2\mathbf x_i \mathbf x_i')\Big[E(\mathbf x_i \mathbf x_i')\Big]^{-1} \qquad [A]$$

with consistent estimator

$$\left(\frac 1n\mathbf X'\mathbf X\right)^{-1}\left(\frac 1n\sum_{i=1}^n\hat u_i^2\mathbf x_i \mathbf x_i'\right)\left(\frac 1n\mathbf X'\mathbf X\right)^{-1} \qquad [\hat A]$$

while the theoretical expression of the asymptotic variance for the usual OLS standard errors is

$$ \sigma^2\Big[E(\mathbf x_i \mathbf x_i')\Big]^{-1}\qquad [B]$$

with consistent estimator

$$ \left(\frac 1n\sum_{i=1}^n\hat u_i^2\right)\left(\frac 1n\mathbf X'\mathbf X\right)^{-1} \qquad [\hat B]$$

Assume that homoskedasticity holds (i.e. that $\sigma^2$ is indeed the constant variance), but we nevertheless estimate and use equation $[\hat A]$ instead of $[\hat B]$. Where is the problem?

The problem is that with eq. $[\hat A]$ we estimate more population parameters than with eq. $[\hat B]$. To see this, write the middle sum in eq. $[\hat A]$ in matrix form. Assume for clarity that the regression is simple, i.e. that we have only a constant term and one stochastic regressor. Then

$$\frac 1n\sum_{i=1}^n\hat u_i^2\mathbf x_i \mathbf x_i' = \left[\begin{matrix} \frac 1n\sum_{i=1}^n\hat u_i^2 & \frac 1n\sum_{i=1}^n\hat u_i^2x_i \\ \frac 1n\sum_{i=1}^n\hat u_i^2x_i & \frac 1n\sum_{i=1}^n\hat u_i^2x_i^2 \\ \end{matrix}\right] \rightarrow_p \left[\begin{matrix} \sigma^2 & \frac 1n\sum_{i=1}^nE(\hat u_i^2x_i) \\ \frac 1n\sum_{i=1}^nE(\hat u_i^2x_i) & \frac 1n\sum_{i=1}^nE(\hat u_i^2x_i^2) \\ \end{matrix}\right]$$

So with both equations we estimate $\sigma^2$ and $E(\mathbf x_i \mathbf x_i')$, but if we use eq. $[\hat A]$ we essentially estimate in addition $\frac 1n\sum_{i=1}^nE(\hat u_i^2x_i)$ and $\frac 1n\sum_{i=1}^nE(\hat u_i^2x_i^2)$, which, if you write them out, include 3d and 4th moments of the regressors (that are "population parameters" also). Now in general, the finite sample properties of an estimator are better the fewer population parameters we have to estimate in order to form/calculate this estimator. Hence, if homoskedasticity holds, we are better off by using eq. $[\hat B]$, since after all, our samples are finite.


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