So it is often advise to use Generalized Least Squares when we have a regression model with non-spherical(i.e. heteroskedastic or autocorrelated) errors. We do so by doing a weighted regression $$ (y-x\hat\beta)^TW(y-x\hat\beta) $$ with $W = \Sigma^{-1} = Cov(\epsilon)^{-1}$, the inverse of the covariance matrix of errors.

The variance of the estimated $\hat\beta$ is $$ \begin{aligned} Var(\hat\beta_{GLS})&=(X^TWX)^{-1}X^TW\Sigma W^TX(X^TW^TX)^{-1}\\ &=(X^TWX)^{-1}=(X^T\Sigma^{-1} W)^{-1} \end{aligned} $$

To do GLS, we must know $\Sigma$. But if we already know $\Sigma$, why can't we just do regular OLS, and calculate $Var(\hat\beta)$ as $$ Var(\hat\beta)=(X^TX)^{-1}X^T\Sigma X(X^TX) $$ ? Is it because $Var(\hat\beta_{GLS})$ is smaller?

Another question I've always had is that for OLS, $\beta$ is estimated as : $$ \hat\beta_{OLS}=(X^TX)^{-1}X^Ty $$ . For GLS or WLS, the $\hat\beta$ is estimated as $$ \hat\beta_{GLS} = (X^TWX)^{-1}X^TWy $$ , which is unbiased for non-spherical error. Yet, we are told that $\hat\beta_{OLS}$ is also an unbiased estimator of $\beta$, even with non-spherical errors. Does that mean $(X^TWX)^{-1}X^TWy$ simplifies to $(X^TX)^{-1}X^Ty$?


1 Answer 1


You can indeed do regular OLS and compute the variance of the estimator (and that estimator will be unbiased and consistent).

But, GLS will be a more efficient estimator that has a lower variance of the sampling distribution (in fact out of all unbiased linear estimators, it will be the estimator with the least possible variance).


Let us estimate $\mu$ with the following variables $$X_k \sim N\left(\mu, \sigma^2 \cdot k\right)$$


$$\begin{array}{} \hat\mu_\text{OLS}& =&\frac{1}{n} \sum_{k=1}^n {X_k} &\sim& N\left(\mu, \sigma^2\cdot{\frac{1+1/n}{2}}\right)\\ \hat\mu_\text{GLS}& =&\frac{1}{H_{n,0.5}}\sum_{k=1}^n \frac{1}{\sqrt{k}} {X_k} &\sim& N\left(\mu, \sigma^2\cdot{\frac{n}{(H_{n,0.5})^2}}\right) \end{array}$$

where $H_{n,0.5} = \sum_{k=1}^n \frac{1}{k^{0.5}}$

The variance of the GLS estimator is smaller than the variance of the OLS estimator. So if you can reasonably guess the covariance matrix of the error distribution $\Sigma$ this may be beneficial.

example comparison OLS and GLS

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    $\begingroup$ That makes sense. What about the estimate of coefficient themselves? Is $\hat\beta_{OLS} = \hat\beta_{GLS}$? If not, how can both be unbiased estimators? $\endgroup$
    – wwyws
    Commented Apr 16, 2022 at 22:16
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    $\begingroup$ GLS estimates $\Sigma$. You just need to tell GLS the assumed correlation pattern (e.g., AR(1)) in order for it to estimate $\Sigma$. $\endgroup$ Commented Apr 16, 2022 at 23:29
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    $\begingroup$ @wwyws - if the estimates were always the same, neither would be better than the other. They can both be unbiased if both their expected values are the same (and equal to the true parameter); with each individual sample, though, their calculated values can still be different, as long as the differences have an expected value equal to zero. $\endgroup$
    – jbowman
    Commented Apr 17, 2022 at 0:13
  • $\begingroup$ @wwyws the example shows how the estimates are different. It is due to the different coefficients in the sum. But because the coefficients both sum up to 1, you get that the estimates have the expectation value $\mu$ and are unbiased. $\endgroup$ Commented Apr 17, 2022 at 7:19
  • $\begingroup$ @Sextus Empiricus - You can not do OLS for heteroscedastic dataset in order to get Best Linear Unbiased Estimation -- it is errorness way for modelling your dependency. $\endgroup$
    – JeeyCi
    Commented Mar 30 at 12:18

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