Curve fitting: What does LAR do?
12 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
able_archer
il 1 Dic 2016
Commentato: able_archer
il 1 Dic 2016
Hello, when conducting a nonlinear least squares regression, one can select 'LAR' (=least absolute residuals) under 'fitoptions'. This is a bit confusing because afaik LAR by definition does NOT minimize squared, but linear residuals. So if 'LAR' is selected, the regression is actually not based on nonlinear least squares, right?
Greetings Vince
1 Commento
Tamir Suliman
il 1 Dic 2016
I think this has been answered in
https://www.mathworks.com/matlabcentral/answers/81143-fit-and-polyfit-lar-vs-least-squares
Risposta accettata
John D'Errico
il 1 Dic 2016
Modificato: John D'Errico
il 1 Dic 2016
I don't understand what is confusing. The sum of the absolute values of the residuals is clearly NOT least squares, thus a minimum sum of squares of residuals. If it was, there would be no point in offering it as an alternative.
Least squares puts more importance on the large residuals. For example, suppose you have two cases, based on two sets of parameters. We need consider only two points.
Case 1: Resid1 = 1, Resid2 = 1
case 2: Resid1 = 0, Resid2 = 2
For a least squares solver, the two cases are very different. Case 1 would be preferred, since the sum of squares is 2, versus 4 for the latter. So whatever set of parameters generated case 1 will be chosen as better.
In a LAR scheme (also called MAD, for minimum absolute deviations) the sum of the absolute residuals are identical. Neither solution would be preferred over the other.
One reason one might use such a scheme is it will be less sensitive to outliers in the data, whereas least squares tends to be jerked around by outliers quite easily.
Più risposte (0)
Vedere anche
Categorie
Scopri di più su Linear and Nonlinear Regression in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!