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

1 Answer 1


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

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