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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!

T34driver
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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.

Gordon Smyth
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  • Thanks. This is very helpful! Do you have any good and detailed reference on the Huber loss or the Huber estimator? – T34driver Nov 08 '21 at 09:40
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    @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. – Gordon Smyth Nov 09 '21 at 01:40
  • 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. – T34driver Nov 09 '21 at 06:39