Timeline for Sum of squares of residuals instead of sum of residuals
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| Jun 11, 2020 at 14:32 | history | edited | CommunityBot |
Commonmark migration
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| Jan 30, 2013 at 18:59 | comment | added | whuber♦ | These comments appear in places to use "variance" in two senses: the variance of the residuals and the variance of the estimates. Among linear estimators (not "all" estimators, pace IMA), least squares minimizes the estimation variance. It is a theorem that the estimation variance is "based on the sum of squares" of the residuals, provided the estimator is linear. @Tomas If the estimator is not linear, then the estimation variance is not proportional to the sum of squares of the residuals, so there's nothing circular about Adam's statement--and he is clear about the assumptions. | |
| Jan 29, 2013 at 13:00 | comment | added | IMA | Gaus-Markov implies that no other method has a smaller variance. If you want to minimize the variance, use least squares. I don't see where it is "going in circles" as much as a thing that makes sense. To complete the answer to the question asked, one would say "We use the squares, instead of the absolutes, because we want to minimize the variance. The GM Theorem shows us that using the squares (doing OLS) is indeed the method which minimizes the variance". It is a perfectly good explanation for using the squares (edit: given all assumptions etc.) | |
| Jan 29, 2013 at 12:54 | comment | added | Adam Bailey | There is a lot of good material on these issues in stats.stackexchange.com/questions/46019/… and stats.stackexchange.com/questions/118/…. | |
| Jan 29, 2013 at 10:55 | comment | added | Adam Bailey | @Tomas Thank you, I see the point, it leads back to the question why, or should, we want estimates of coefficients to be precise as measured by minimum variance, rather than some other measure of precision. Having said that, minimum variance is one popular measure of precision, so the G-M Theorem helps to explain why OLS regression is widely used. | |
| Jan 29, 2013 at 10:18 | comment | added | Tomas | "More efficient implies that the variances are lower" - I think you are going in circles, because variance is based on the sum of squares. Had you been using some other measure based on absolute values instead, it would possibly favor the absolute values. | |
| Jan 29, 2013 at 9:41 | history | edited | Adam Bailey | CC BY-SA 3.0 |
added 482 characters in body
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| Jan 28, 2013 at 19:13 | history | answered | Adam Bailey | CC BY-SA 3.0 |