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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? $\endgroup$ – Christoph Hanck May 14 '18 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 '18 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 '18 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. $\endgroup$ – Michael Chernick May 14 '18 at 20:48
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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 '18 at 20:49

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