[R] Robust regression for ordered data

David Winsemius dwinsemius at comcast.net
Tue Oct 9 02:25:34 CEST 2012


On Oct 8, 2012, at 5:17 PM, Eiko Fried wrote:

>> It **might** also help the OP clarify his intent in
>> exactly the way you describe (or not).
> 
> 
> My intent is to perform a linear regression on a metric dependent variable.

What do you mean by "a metric dependent variable"?

> Unfortunately, two key assumptions - normal distribution and
> homoscedasticity - cannot be met.

You have not demonstrated that these "key assumptions" have not been met. Those assumptions refer to the residuals after the model is fit. Many novice practitioners incorrectly assume they refer to the predictor variables.

> 
> I read that using a robust regression is the way to go in this case, and I
> wanted to ask whether this is possible in R, and if so, whether there is a
> robust regression that can handle the lack of both assumptions.
> 
> On top of that, I get different results using the Huber and Bisquare
> weighting functions, and don't know what that implies.

Nor do we, ... since you have shown neither that data nor the code.

-- 
David.
> 
> Thanks
> 
> 
> 
>> 
>> On Sun, Oct 7, 2012 at 10:47 PM, Prof Brian Ripley
>> <ripley at stats.ox.ac.uk> wrote:
>>> On 08/10/2012 00:37, Bert Gunter wrote:
>>>> 
>>>> Have you checked the Robust task view on CRAN?? Would seem that that
>>>> should have been the first place to look.
>>> 
>>> 
>>> It is still a conceptual question.  I presume this means an ordered
>>> response, and then we need to know what is meant by 'regression'.
>>> 
>>> If you tell us precisely what robust method you want to know about, you
>> may
>>> get help about whether it is available in R.   But I surmise that you
>> need
>>> rather to be looking at ordinal regression (polr in MASS, for example),
>> and
>>> you will not find that in the 'Robust' task view.   In the task view,
>>> 'robust' is  a technical term and I don't think 'Elko Fried' is using it
>> in
>>> the sense the author of the task view is.
>>> 
>>>> 
>>>> -- Bert
>>>> 
>>>> On Sun, Oct 7, 2012 at 3:30 PM, Eiko Fried <torvon at gmail.com> wrote:
>>>>> 
>>>>> Thank you Jeff! Please ignore the first of my two questions then, and
>>>>> apologies for not making it clear that my second question was about R.
>>>>> 
>>>>> (2) "Are there ways of using robust regressions with ordered data" ...
>> in
>>>>> R?
>>>>> 
>>>>> Thank you
>>>>> 
>>>>> 
>>>>> On 7 October 2012 18:26, Jeff Newmiller <jdnewmil at dcn.davis.ca.us>
>> wrote:
>>>>> 
>>>>>> This does not appear to be a question about R. You should post in a
>> list
>>>>>> or forum dedicated to discussing statistics theory, such as
>>>>>> stats.stackoverflow.com.
>>>>>> 
>>>>>> 
>> ---------------------------------------------------------------------------
>>>>>> Jeff Newmiller                        The     .....       .....  Go
>>>>>> Live...
>>>>>> DCN:<jdnewmil at dcn.davis.ca.us>        Basics: ##.#.       ##.#.  Live
>>>>>> Go...
>>>>>>                                       Live:   OO#.. Dead: OO#..
>>>>>> Playing
>>>>>> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
>>>>>> /Software/Embedded Controllers)               .OO#.       .OO#.
>>>>>> rocks...1k
>>>>>> 
>>>>>> 
>> ---------------------------------------------------------------------------
>>>>>> Sent from my phone. Please excuse my brevity.
>>>>>> 
>>>>>> Eiko Fried <torvon at gmail.com> wrote:
>>>>>> 
>>>>>>> I have two regressions to perform - one with a metric DV (-3 to 3),
>> the
>>>>>>> other with an ordered DV (0,1,2,3).
>>>>>>> 
>>>>>>> Neither normal distribution not homoscedasticity is given. I have a
>> two
>>>>>>> questions:
>>>>>>> 
>>>>>>> (1) Some sources say robust regression take care of both lack of
>> normal
>>>>>>> distribution and heteroscedasticity, while others say only of normal
>>>>>>> distribution. What is true?
>>>>>>> (2) Are there ways of using robust regressions with ordered data, or
>> is
>>>>>>> that only possible for metric DVs?
>>>>>>> 
>>>>>>> Thanks
>>>>>>> Torvon
>>>>>>> 


David Winsemius, MD
Alameda, CA, USA




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