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I have build a $\rho$ function which has the following definition:

\begin{equation} \rho(x)= \left\{ \begin{array}{ll} 4- \frac{8}{x^2} \text{if } x \lt-2\\ \frac{x^2}{2} \text{if } x \in [-2,3] \\ 9-\frac{81}{2*x^2} \text{if } x \gt 3 \end{array} \right. \end{equation}

The function has the continuity and derivability properties, but is asymetric. Can be solved using Iteratively reweighted least squares (IRLS) method?

The $u$ function is:

$ u(x)=\begin{equation} \left\{ \begin{array}{ll} \frac{16}{x^4} \text{ if } x \lt-2\\ 1 \text{ if } x \in [-2,3] \\ \frac{81}{x^4} \text{ if } x \gt 3 \end{array} \right. \end{equation} $

For the IRLS method, the symmetry of $u$ function is mandatory?

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No [question mark in the title]: only the re-descending behavior matters from a bdp point of view. Yes [first question mark in the text], from an IRLS point of view this loss function is not more difficult than, say, Huber's. No [second question mark], there are other asymmetric $M$ functions and these could be solved by IRLS (although in practice they are not). But i'd be remiss if i didn't ask: what are you trying to achieve? – user603 Jul 17 '12 at 9:56
I need my estimator instead of Huber M estimator. The solution is to use $u$ function in IRLS method. I don't know what is the effect of symmetry property. – Ion Caciula Jul 17 '12 at 19:17

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