It seems to me that since $$ \widehat{\vec{\theta}} = \mathrm{argmin}_{\vec{\theta}} \sum_{i=1}^{n} - \log(f(x_i; \vec{\theta})) = \sum_{i=1}^{n} - \log\left( \frac{\partial F}{\partial x}(x_i; \vec{\theta} )\right) = \widehat{\vec{\theta}}(F), $$ that this means that maximum likelihood estimators are robust estimators. Is this correct?

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    $\begingroup$ Robust in what sense? $\endgroup$ May 14, 2018 at 18:31
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    $\begingroup$ Thank you for your comment, what could be confusion be? I thought M-estimators were robust estimators, and M-estimators were generalizations of MLE estimators? $\endgroup$
    – Mikkel Rev
    May 14, 2018 at 18:36
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    $\begingroup$ Why did you think M-estimators are robust? The sample mean is an M-estimator, and it is famously non-robust. $\endgroup$
    – jbowman
    May 14, 2018 at 18:57
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    $\begingroup$ The fact that the MLE maximizes the likelihood indicates that it is dependent on the assumed family of distributions and so would not be good estimates when the model departs greatly from the underlying assumed family. $\endgroup$ May 14, 2018 at 20:48

2 Answers 2


By the definition of robust estimators, this is true. That is, M-estimators are a type of robust statistics, and MLE's are a special case of M-estimators.

However, it's definitely not the case that MLE's in general have good robustness properties.

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    $\begingroup$ from that link: 'There are various definitions of a "robust statistic."' Which definition? $\endgroup$
    – Taylor
    May 14, 2018 at 20:49

One way to look at it is that there is no strict distinction between robust and non-robust estimators, but rather they can be compared according to their robustness properties (as already pointed in the answer by @CliffAB.) Thus, M-estimators can be viewed as explicitly designed to resemble the maximum likelihood estimators (see, e.g., their introduction in Robust Statistics: Theory and Methods by Maronna, Martin and Yohai.). That is, M-estimators are defined as minima of a function of the type shown in the OP (or as zeros of a derivative of such a function), which however does not necessarily have an interpretation of likelihood (e.g., the associated probability density $f(x; \vec{\theta})$ might not be normalizable.)

Thus, both sample mean and sample median can be viewed as M-estimators for location, but the former with breakdown point of 0 and the latter with breakdown point of 50%.

This suggests the other way to look at: by defining the non-robust estimators as those with extremely poor measures of robustness. Thus, we could define as non-robust any estimator with breakdown point of 0, meaning that even a single outlying observation may produce arbitrarily big deviation from the true value of the parameter being estimated.


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