Reputation
Next privilege 125 Rep.
Vote down
Badges
3
Newest
 Editor
Impact
0 people reached

  • 0 posts edited
  • 0 helpful flags
  • 129 votes cast
Jul
7
comment Fitting - Exponential
@MichaelSeifert I have held the same impression myself and god only knows how many such analyses I have done. But if you ponder for a moment you will see that weighted NLLS and unweighted LLS cannot be equal. In the weighted case, you are fitting (wx,wy) where w includes weights. In the nonweighted case, you are fitting (log(x),log(y)). The difference is that w is a constant and it is the SAME on both x and y. Taking the log is equivalent to different "weights" on both sides unless x and y are identical (in which case the fitting is unnecessary because y=x will fit the data).
Jul
7
comment Fitting - Exponential
@MarcelKiesel Please see my comment below. In addition to using the linear LS estimates as a starting point for the nonlinear LS fit, I would also use some weights. It seems like you care a lot about accuracy, especially with the smaller points so simply use weighted LS and give the smaller points more weight. Do you have any idea of the variance/errors in these measurements? The ideal weights would be something like 1/variance = 1/error^2 so points with smaller errors get more weight because they are more accurate and usually smaller magnitudes mean smaller errors.
Jul
7
comment Fitting - Exponential
@MichaelSeifert I would argue here that doing a nonlinear least squares fit on the original data is not the same as linear least squares fit in log-log space. If you assume that the original data satisfies the assumptions of least squares then taking the log of the data will violate those assumptions and LS will give wrong/biased results. One could take the linear LS estimates and use those as initial conditions for the correct nonlinear LS fit.
Sep
26
awarded  Editor
Mar
10
awarded  Teacher
Dec
26
awarded  Supporter