For a linear model the OLS solution provides the best linear unbiased estimator for the parameters.

Of course we can trade in a bias for lower variance, e.g. ridge regression. But my question is regarding having no bias. Are there any other estimators that are somewhat commonly used, which are unbiased but with a higher variance than the OLS estimated parameters?

If I would have a huge data set I could of course sub-sample it and estimate the parameters with less data, and increase the variance. I assume this could be hypothetically useful.

This is more of a rhetorical question, because when I have read about BLUE estimators, a worse alternative is not provided. I guess that providing worse alternatives could also help people understand the power of BLUE estimators better.

  • $\begingroup$ What about a maximum likelihood estimator? E.g. if you think your data is sampled from a $t$ distribution with a relatively low degrees of freedom parameter ($t(3)$ or $t(4)$ may be characteristic to financial returns), a maximum likelihood estimator would not coincide with OLS but I guess it would still be unbiased. $\endgroup$ Commented May 21, 2015 at 17:44
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    $\begingroup$ Relevant: andrewgelman.com/2015/05/11/… $\endgroup$ Commented May 22, 2015 at 13:31
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    $\begingroup$ @RichardHardy, I also tried the MLE, with the results you anticipated. $\endgroup$ Commented Mar 11, 2019 at 9:47

1 Answer 1


One example that comes to mind is some GLS estimator that weights observations differently although that is not necessary when the Gauss-Markov assumptions are met (which the statistician may not know to be the case and hence apply still apply GLS).

Consider the case of a regression of $y_i$, $i=1,\ldots,n$ on a constant for illustration (readily generalizes to general GLS estimators). Here, $\{y_i\}$ is assumed to be a random sample from a population with mean $\mu$ and variance $\sigma^2$.

Then, we know that OLS is just $\hat\beta=\bar y$, the sample mean. To emphasize the point that each observation is weighted with weight $1/n$, write this as $$ \hat\beta=\sum_{i=1}^n\frac{1}{n}y_i. $$ It is well-known that $Var(\hat\beta)=\sigma^2/n$.

Now, consider another estimator which can be written as $$ \tilde\beta=\sum_{i=1}^nw_iy_i, $$ where the weights are such that $\sum_iw_i=1$. This ensures that the estimator is unbiased, as $$ E\left(\sum_{i=1}^nw_iy_i\right)=\sum_{i=1}^nw_iE(y_i)=\sum_{i=1}^nw_i\mu=\mu. $$ Its variance will exceed that of OLS unless $w_i=1/n$ for all $i$ (in which case it will of course reduce to OLS), which can for instance be shown via a Lagrangian:

\begin{align*} L&=V(\tilde\beta)-\lambda\left(\sum_iw_i-1\right)\\ &=\sum_iw_i^2\sigma^2-\lambda\left(\sum_iw_i-1\right), \end{align*} with partial derivatives w.r.t. $w_i$ set to zero being equal to $2\sigma^2w_i-\lambda=0$ for all $i$, and $\partial L/\partial\lambda=0$ equaling $\sum_iw_i-1=0$. Solving the first set of derivatives for $\lambda$ and equating them yields $w_i=w_j$, which implies $w_i=1/n$ minimizes the variance, by the requirement that the weights sum to one.

Here is a graphical illustration from a little simulation, created with the code below:

EDIT: In response to @kjetilbhalvorsen's and @RichardHardy's suggestions I also include the median of the $y_i$, the MLE of the location parameter pf a t(4) distribution (I get warnings that In log(s) : NaNs produced that I did not check further) and Huber's estimator in the plot.

enter image description here

We observe that all estimators seem to be unbiased. However, the estimator that uses weights $w_i=(1\pm\epsilon)/n$ as weights for either half of the sample is more variable, as are the median, the MLE of the t-distribution and Huber's estimator (the latter only slightly so, see also here).

That the latter three are outperformed by the OLS solution is not immediately implied by the BLUE property (at least not to me), as it is not obvious if they are linear estimators (nor do I know if the MLE and Huber are unbiased).

n <- 100      
reps <- 1e6

epsilon <- 0.5
w <- c(rep((1+epsilon)/n,n/2),rep((1-epsilon)/n,n/2))

ols <- weightedestimator <- lad <- mle.t4 <- huberest <- rep(NA,reps)

for (i in 1:reps)
  y <- rnorm(n)
  ols[i] <- mean(y)
  weightedestimator[i] <- crossprod(w,y)  
  lad[i] <- median(y)   
  mle.t4[i] <- fitdistr(y, "t", df=4)$estimate[1]
  huberest[i] <- huber(y)$mu

plot(density(ols), col="purple", lwd=3, main="Kernel-estimate of density of OLS and other estimators",xlab="")
lines(density(weightedestimator), col="lightblue2", lwd=3)     
lines(density(lad), col="salmon", lwd=3)     
lines(density(mle.t4), col="green", lwd=3)
lines(density(huberest), col="#949413", lwd=3)
legend('topright', c("OLS","weighted","median", "MLE t, 4 df", "Huber"), col=c("purple","lightblue","salmon","green", "#949413"), lwd=3)
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    $\begingroup$ Neat! I think this is a very simple illustrative example, bit more general than the one I came up with. When people are learning about estimators in a frequentist setting I feel that these kind of examples are often missing, they really help you get a better grasp of the concept. $\endgroup$
    – Gumeo
    Commented May 21, 2015 at 12:46
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    $\begingroup$ Another possibility would be (robust) estimators based on minimizing a criterion such as $W=\sum_{i=1}^n w(e_i)$ where $e_i$ is the ith residual and $w$ is some symmetric function, convex or non-convex, with (global) minimum at 0, $w(0)=0$. The Huber estimator would be an example. $\endgroup$ Commented May 22, 2015 at 13:23
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    $\begingroup$ @kjetilbhalvorsen, I now also include the Huber estimator, which actually does rather well. $\endgroup$ Commented Mar 11, 2019 at 10:44

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