What location parameter is modelled by robust regression? There is quite some number of ways how to robustly fit a linear regression model, e.g. using M-estimation based on Tukey's biweight loss or on Huber's loss, see e.g. Wikipedia.
I got two questions about the two mentioned cases:


*

*Is it correct that these are modelling the mean (more precisely the center) of the conditional response distribution as long as the error distribution is symmetric?

*What location parameter are they modelling in general? Is it a specially weighted average of the response variable?

 A: 
There is quite some number of ways how to robustly fit a linear regression model, e.g. using M-estimation based on Tukey's biweight loss or on Huber's loss, see e.g. Wikipedia.
Is it correct that these are modelling the mean of the response as long as the error distribution is symmetric?

Well, estimating, but that may be what you mean by modelling -- but only if the mean exists. But they'll be reasonable estimates of the center of symmetry more generally, which will be the population median, and trimean, and pseudomedian, and any trimmed mean, and, ... and also usually the mean. For example, consider the $t$ distribution with $\nu$ degrees of freedom. If $\nu\leq 1$ there's no mean. But it still has a center of symmetry.

What location parameter are they modelling in general?

M-estimation corresponds to maximum-likelihood estimation if the loss function corresponds to $-\log(f)$  for some density $f$. This is the case for the Huber loss but not the Tukey.

Is it a specially weighted average of the response variable?

Not in general no*. But M-estimators can be obtained** by iterating a weighted average where the weights are updated at each step.
*(at least not unless 'special' is interpreted much more broadly than people generally understand the term 'weighted average' to be able to stretch)
**(in some circumstances at least)
