# Are maximum likelihood estimator robust estimators?

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?

• Robust in what sense? May 14, 2018 at 18:31
• 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? May 14, 2018 at 18:36
• Why did you think M-estimators are robust? The sample mean is an M-estimator, and it is famously non-robust. May 14, 2018 at 18:57
• 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. May 14, 2018 at 20:48

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.)