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Assume, I have a list of actual, noisy, independent scalar measurements

  • [y1, y2, ..., yn] (think: y1 = [0%, 12%, 52%, 79%, 99%]) for a scalar series
  • x (think: x = [0 min, 2 min, 4 min, 6 min, 10 min]).

under the same conditions.

I know the underlying function y= f(x, a, b, c) and want to determine the coefficients a, b, c with curve fitting (least-square). What is the theoretically correct approach?

  • A) Determine a1, b1, c1 for (y1, x) etc. and calculate a as the mean of a1, a2, ..., an, the same for b and c?
  • B) Determine a, b, c in a curve fit for all pairs [(y1, x), (y2, x), ..., (yn, x)] simultaneously?

I was sure that A) is the correct approach but in retrospect and having spent some time on SciPy's curve fitting, I am not so sure that this was not a practical limitation because the commercially available software used only allowed A). However, in B), one would lose the connection within each independent observation, y1, y2, ..., yn.

I would be surprised if there were no duplicate questions but I cannot find them, so any comment is welcome.

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(B) should find the set of parameters a,b, c that minimizes the overall sum of squared errors (i.e. the overall squared error across all "time points" 1,2,...,n). You don't have the same guarantee if you estimate the parameter separately and then average as you suggest in (A).

Imho (A) instead seems justified only if there's reasons to expect that the "true" parameters take different value at each point 1,2,...,n. However in this case you wouldn't want to average them.

By the way, if your measurement y is a percentage, bounded within 0% and 100%, least squares might not be the best estimation method. There are alternative approaches designed to work with bounded variables, for example have a look at beta-regression.

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  • $\begingroup$ No, the coefficients should not be different for the measurements y1, ..., yn. These y-series data are just repetitions of the same experiment under the same condition. Your answer reinforces my understanding that the software forced a specific stats model on my mind. $\endgroup$ – Mr. T Dec 21 '20 at 18:57

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