I busy with a regression model that seems to have heteroscedasticity. The model has 6 independent variables and one dependent variable. I did the regression and noticed heteroscedasticity. I then eliminated the observations with very high residuals +/- 5% of the total observations which is 25000. I did the regression again without these high residual observations and found that this new regression does not have any signs of heteroscedasticity. I would like to know: is a process like this acceptable or not?
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asked Feb 22, 2015 at 14:35
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$\begingroup$ "I busy ...": I haven't edited what I don't understand: perhaps dialect or slang beyond my region or age group. "+/-" presumably means about. Did you delete observations with just large positive residuals, large positive and large negative, etc.? $\endgroup$– Nick CoxFeb 22, 2015 at 14:56
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$\begingroup$ Everything is acceptable, it just depends to whom;). When you say regression, you mean plain vanilla OLS? $\endgroup$– user603Feb 22, 2015 at 17:55
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You ask for a yes-no answer, but answers might vary greatly. I won't be the only member here unwilling to approve (or condemn) what I can't see.
From this description I'd say that the answer might vary from
- that was brutal and utterly ad hoc, but it is just possible that you got a fair model with poor methods
through
- we can't tell, because just mentioning heteroscedasticity does give any context to judge what is a good model; we need to see the data and know what model you tried (there is zero formal content in your post, although if I had to guess it's linear regression on the variables as given with no extra complications)
to
- no; it's unacceptably poor practice to drop observations because they happen to be inconvenient compared with simplistic assumptions that would be nice if true.
Personally, I go with all of these, but my bottom line is as just given.
On this information, I would suggest some priorities:
Heteroscedasticity is arguably a secondary problem to do with error structure; you may need much more emphasis on getting the functional form and estimation method right.
Additionally, using transformations or a non-identity link function is a much better way to proceed than dropping some fraction of the data.
I'd add that as your analysis was demonstrably two-phase, the resulting standard errors and P-values from the second leg don't mean much and should certainly not be reported as if the dataset arrived as you left it.
answered Feb 22, 2015 at 14:53
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$\begingroup$ As a counter example to the dangers of unwarranted transforming autobox.com/pdfs/vegas_ibf_09a.pdf slides 14-24 . Transformations should be based upon the distribution of the residuals from a model not on the distribution of the original series because parametric tests require a central ch-square distribution. Another possible translation for busy is "am currently working with" $\endgroup$ Feb 22, 2015 at 15:35
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$\begingroup$ No one approves of unwarranted transformation, but I'd say your dictum (Diktat?) on transforming because of how residuals behave is contestable. If the pattern is (just one example) $y = a \exp(bx)$ it can make sense to work with $\ln y$, almost regardless of the exact pattern of residuals. $\endgroup$– Nick CoxFeb 22, 2015 at 16:16
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$\begingroup$ The rubber hits the road when you try and use standard parametric tests of significance and sufficiency when the residuals are non-normal. $\endgroup$ Feb 22, 2015 at 18:54
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$\begingroup$ I don't how you test sufficiency; it's a mathematical property, not the subject of hypotheses. I guess I feel more comfortable than @IrishStat in working with distributions other than normal; we have had most of the useful ones for decades. Minor sparring apart, the process in this Q is a count process, so it's not especially clear that forcing such counts into a normal straitjacket would be a good idea. $\endgroup$– Nick CoxFeb 22, 2015 at 19:23
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$\begingroup$ The idea of model sufficiency is that if there is identifiable information in the residuals then there is an argument for adding/moving structure to the model whereas necessity is the concept of do I need the structure I have. $\endgroup$ Feb 22, 2015 at 19:35