I have read in the abstract of this paper that:

"The maximum likelihood (ML) procedure of Hartley aud Rao is modified by adapting a transformation from Patterson and Thompson which partitions the likelihood render normality into two parts, one being free of the fixed effects. Maximizing this part yields what are called restricted maximum likelihood (REML) estimators."

I also read in the abstract of this paper that REML:

"takes into account the loss in degrees of freedom resulting from estimating fixed effects."

Sadly I don't have access to the full text of those papers (and probably would not understand if I did).

Also, what are the advantages of REML vs. ML? Under what circumstances may REML be preferred over ML (or vice versa) when fitting a mixed effects model? Please give an explanation suitable for someone with a high-school (or just beyond) mathematics background!


3 Answers 3


As per ocram's answer, ML is biased for the estimation of variance components. But observe that the bias gets smaller for larger sample sizes. Hence in answer to your questions "...what are the advantages of REML vs ML ? Under what circumstances may REML be preferred over ML (or vice versa) when fitting a mixed effects model ?", for small sample sizes REML is preferred. However, likelihood ratio tests for REML require exactly the same fixed effects specification in both models. So, to compare models with different fixed effects (a common scenario) with an LR test, ML must be used.

REML takes account of the number of (fixed effects) parameters estimated, losing 1 degree of freedom for each. This is achieved by applying ML to the least squares residuals, which are independent of the fixed effects.

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    $\begingroup$ Indeed, the REML estimator of a variance component is usually (approximately) unbiased, while the ML estimator is negatively biased. However, the ML estimator usually has lower mean-squared error (MSE) than the REML estimator. So, if you want to be right on average, go with REML, but you pay for this with larger variability in the estimates. If you want to be closer to the true value on average, go with ML, but you pay for this with negative bias. $\endgroup$
    – Wolfgang
    Commented Apr 3, 2014 at 8:44
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    $\begingroup$ In the simple case of a constant mean, and constant variance, ML is dividing SSR with $n$ while REML is dividing SSR with $(n-1)$. So REML is a generalization of this procedure! $\endgroup$ Commented May 4, 2015 at 16:58
  • $\begingroup$ "ML is biased for the estimation of variance components". Does it mean the variance of the random effects or also the the standard errors of the fixed-effect coefficientes? $\endgroup$
    – skan
    Commented Jan 17, 2019 at 16:07

Here is a quick answer...

Standard illustrative example

Let $y = (y_1, \dotsc, y_n)$ be a sample from a normal distribution $\mathrm{N}(\mu, \sigma^2$). Both $\mu$ and $\sigma^2$ are unknown. The maximum likelihood estimator of $\sigma^2$, obtained by taking the derivative of the log-likelihood with respect to $\sigma^2$ and equating to zero, is $$ \hat{\sigma}^2_{\textrm{ML}} = \frac{1}{n} \sum_{i=1}^n (y_i -\bar{y})^2 $$ where $\bar{y} = \frac{1}{n} \sum_{i=1}^n y_i$ is the maximum likelihood estimator of $\mu$. We can show that $$ \mathrm{E}(\hat{\sigma}^2_{\textrm{ML}}) = \frac{n-1}{n} \sigma^2. $$ [Start by rewriting $\hat{\sigma}^2_{\textrm{ML}}$ as $\frac{1}{n} \sum_{i=1}^n \left((y_i - \mu) + (\mu - \bar{y})\right)^2$]. Thus, $\hat{\sigma}^2_{\textrm{ML}}$ is biased. Note that if we had known $\mu$, then the MLE for $\sigma^2$ would have been unbiased. Hence, the problem with $\hat{\sigma}^2_{\textrm{ML}}$ appears to be linked with the fact that we have substituted $\bar{x}$ for the unknown mean in the estimation. The intuitive idea of REML estimation is to end up with a likelihood that contains all the information on $\sigma^2$ but no longer contains the information on $\mu$.

More technically, the REML likelihood is a likelihood of linear combinations of the original data: instead of the likelihood of $y$, we consider the likelihood of $Ky$, where the matrix $K$ is such that $\mathrm{E}[Ky] = 0$.

REML estimation is often used in the more complicated context of mixed models. Every book on mixed models have a section explaining REML estimation in more details.


@Joe King: Here (and here, for English web page context) is one of my favorite books on mixed models that is fully available online. Section 2.4.2 deals with estimating variance components. Enjoy your reading :-)

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    $\begingroup$ Thank you - this is helpful - though I don't have easy access to books on mixed models. Please could you relate your answer to the 2 quotes in my post ? $\endgroup$
    – Joe King
    Commented Jan 28, 2013 at 18:17
  • $\begingroup$ I wonder how a multivariate Gaussian changes the story? stats.stackexchange.com/questions/167494/… $\endgroup$ Commented Aug 18, 2015 at 1:19

ML method underestimates the variance parameters because it assumes that the fixed parameters are known without uncertainty when estimating the variance parameters.

The REML method uses a mathematical trick to make the estimates for the variance parameters independent of the estimates for the fixed effects. REML works by first getting regression residuals for the observations modeled by the fixed effects portion of the model, ignoring at this point any variance components.

ML estimates are unbiased for the fixed effects but biased for the random effects, whereas the REML estimates are biased for the fixed effects and unbiased for the random effects.


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