# Correct standard errors for weighted linear regression

What is the correct way to calculate the standard errors of the coefficients in a weighted linear regression?

The regression equation I am using is $y_i = a + bx_i$, and I have weights, $w_i = 1/\sigma_i$. The numerical recipes formula for a straight line fit, and the formula given in "An introduction to error analysis" by J. R Taylor, (and Wikipedia too) state that the standard error in the $b$ coefficient is calculated as $$\sigma_b = \sqrt{\frac{\sum w_i}{\sum w_i\sum w_i x_i^2-(\sum w_i x_i)^2}}$$ (or alternatively in matrix form the standard errors are, $\sigma^2 = (X'WX)^{-1}$). This formula can be derived from propagation of errors.

Using R's $lm()$ function (and python's StatsModels), I get a standard error in the $b$ coefficient which appears* to be calculated as $$\sigma_b = \sigma_e\sqrt{\frac{\sum w_i}{\sum w_i\sum w_i x_i^2-(\sum w_i x_i)^2}}$$ where $\sigma_e^2 = \sum w_i(y_i - a - bx_i)^2/(N-2)$ (alternatively, $\sigma^2 = \sigma_e^2(X'WX)^{-1}$ ). So they are the same, except for the $\sigma_e$ multiplier in R and StatsModel.

Is it possible that these actually different measures that are just being called the same thing? Is one preferred over the other for an estimate of the standard error?

*I say "appears" because I couldn't find the actual formula anywhere.

edited because I had omitted the weight terms in the denominators.

• Units analysis provides some insight. Weights are unitless, whence your first formula has units that are the reciprocals of the units of the $x_i$. The second formula (after you fix the errors in your formulas for $\sigma_e$) will have units that are ratios of those of the $y_i$ to those of the $x_i$. Which of those actually is a slope?
– whuber
Feb 23, 2015 at 22:29
• Thanks @whuber. I have corrected the $\sigma_e$ formula and the denominator terms as I and forgotten the weight terms. This was just a typo and doesn't change my question. Feb 24, 2015 at 14:33
• Following on, I get what you mean with regards to the units. If I understand correctly, in the case that the weights are not dimensionless but actually represent the variance of the $y_i$ measurements then the first equation would be the one to use. Whereas if they are dimensionless, then the second equation would be used as $\sigma_e$ is then required to provide an estimate the variance. Feb 24, 2015 at 14:42
• That's almost right. The first equation would make sense only when the weights are reciprocals of variances of the $y_i$.
– whuber
Feb 24, 2015 at 17:12
• The first approach is very common in some engineering and physics related fields, while in other fields we only assume that the weights are defined as relative weights. Some statistical packages provide both, statsmodels will have an option to fix the scale for WLS in the upcoming release 0.7. Mar 1, 2015 at 4:01

These two expressions disagree, as you note, in terms of the use of the residuals in calculation: the difference of $Y$ from the predicted values, being included or omitted from the calculation of the standard errors. They are indeed different estimators but they converge to the same thing in the long run. They also can be combined to create a "sandwich" estimator.

To revisit some basic modeling assumptions: the weighted linear regression model is estimated from a weighted estimating equation of the form:

$$U(\beta) = \mathbf{X}^T \mathbf{W}\left( Y - \mathbf{X}^T\beta\right)$$

When $\mathbf{W}$ is just the diagonal matrix of weights. This estimating equation is also the normal equations (partial log likelihood) for the MLE. Then, the expected information is:

$$\mathbf{A}= \frac{\partial U(\beta)}{\partial \beta} = \mathbf{X}^T\mathbf{W} \mathbf{X}$$

Then $\mathbf{A}^{-1}$ is a consistent estimator of the covariance matrix for $\beta$ when 1. the mean model is appropriately specified and 2. the weights are the inverse variance of the residuals. You have already stated the A matrix is your first display. Contrast this with the observed information:

$$\mathbf{B} = E[U(\beta)U(\beta)^T] = \mathbf{X}^T \mathbf{W}E((Y-\mathbf{X}\beta)^T(Y-\mathbf{X}\beta)) \mathbf{W}\mathbf{X}$$

One of the weight matrices can multiply with the squared errors and factor out of the expression as a constant because it is orthogonal to the $\mathbf{X}$, and you'll note that is the expression for $\sigma_e^2= \sum_{i=1}^n w_i (y_i - a - bx_i)/(n-2)$. $\mathbf{B}$ is also a consistent estimator of the information matrix, but will disagree with $\mathbf{A}$ in finite samples.

As for which one to use, why not use both? A sandwich estimator is obtained by $\left(\mathbf{A}^T\mathbf{B}\mathbf{A}\right)^{-1}$ and depends neither on the mean model being correct nor on the weights being properly specified.