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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)

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)

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. 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)

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)

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, at least -- but only if the mean exists. But they'll be reasonable estimates of the center of symmetry more generally. For example, consider the $t$ distribution with $\nu$ degrees of freedom. If $\nu\leq 1$ there's no mean.

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')

**(in some circumstances at least)

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. 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)