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In Casella and Berger (page484), the following Huber loss is defined. enter image description here

Then on the next page, Table 10.2.1 shows the Huber estimator for different $k$:

enter image description here

In particular, $k=0$ gives the median, which doesn't make any sense. As when $k=0$, the Huber loss according to (10.2.2) is identically 0. But in order to get the median the huber loss has to be the absolute value (possibly with a shift). I guess the form of Huber loss that he gives here is wrong. What's the correct form of Huber loss then?

Thanks!

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Equation 10.2.2 in Casella & Berger is correct. You are right of course that $\rho(x)$ is identically zero when $k=0$ but the estimate is nevertheless defined at $k=0$ in a limiting sense. The value of $a$ that minimizes the criterion function converges to the median as $k$ converges to zero from above. The estimate can be made arbitrarily close to the median by setting $k$ to a sufficiently small positive value.

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  • $\begingroup$ Thanks. This is very helpful! Do you have any good and detailed reference on the Huber loss or the Huber estimator? $\endgroup$
    – ExcitedSnail
    Commented Nov 8, 2021 at 9:40
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    $\begingroup$ @T34driver You don't consider Casella & Berger or Huber (1964) to be good references? One of the Huber's books would be my first thought. There are many Huber estimators. A Google Scholar search for "Huber loss" (with quotes) brings up over 6000 publications, but they are mostly written for a specialist mathematical research audience. $\endgroup$
    – Gordon Smyth
    Commented Nov 9, 2021 at 1:40
  • $\begingroup$ Thanks a lot, professor. Casella&Berger is the best book on statistical inference I have ever read of O(∩_∩)O, I was thinking about reference that is more specialized on Huber-type estimation. Will definitely check out Huber(1964). Many thanks again. $\endgroup$
    – ExcitedSnail
    Commented Nov 9, 2021 at 6:39

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