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This is a question on non-linear curve fitting, not on model selection and not particularly on how to do regression in matlab.

I have 2 models for the same physical process, each a function of the vector x, and both have two coefficients, for m1, they are x0 and b and for m2 they are x0 and theta, where theta is notationally, b/x0. The models return slightly different values for x0 and theta from m2 does not general equal b/x0 from m1.

I am looking for advice on how to justify the use of one model over the other.

x = linspace(0,1,1e3); % For example

m1 = @(x) (1 - (x./x0)) .* exp(-b .* (x/x0));

m2 = @(x) (1 - (x./x0)) .* exp(-theta .* x);

Thanks

Collin

the cyclist
on 5 May 2021

It seems to me that your two models are equivalent, but one is specified in a way that masks that fact.

x0 is really just a scaling parameter -- you can think of it as the "units" of x. Define a new variable:

z = x/x0;

Then

mz1 = @(z) (1 - z) .* exp(-b .* z);

mz2 = @(z) (1 - z) .* exp(-theta .* x0 .* z);

and then define a new parameter

b2 = theta .* x0;

resulting in

mz1 = @(z) (1 - z) .* exp(-b .* z);

mz2 = @(z) (1 - z) .* exp(-b2 .* z);

and you see that your models are equivalent, 1-parameter models.

the cyclist
on 6 May 2021

Actually, I have to take back what I said about these boiling down to a 1-parameter model. Because you don't know x0 a priori, you can't do that step.

However, the two models are certainly equivalent.

% Set random number generator seed, for reproducibility

rng default

% Generative process for creating the data, with a little noise in y.

b = 2;

x0 = 3;

theta = 5;

m1 = @(x) (1 - (x./x0)) .* exp(-b .* (x/x0));

x = linspace(0,1,20)';

y = m1(x) + 0.02*randn(size(x));

% The two forms of the fitting function

f1 = @(B1,x) (1 - (x./B1(1))) .* exp(-B1(2) .* (x./B1(1)));

f2 = @(B2,x) (1 - (x./B2(1))) .* exp(-B2(2) .* x);

% The two fits

mdl1 = fitnlm(x,y,f1,[1 1])

mdl2 = fitnlm(x,y,f2,[1 1])

% The predicted y values from the fits

y1 = predict(mdl1,x);

y2 = predict(mdl2,x);

% Plot the data and fits

figure

hold on

scatter(x,y)

plot(x,y1,'-')

plot(x,y2,'--')

legend({'data','1st model','2nd model'})

% Compare coefficents

coef1 = mdl1.Coefficients.Estimate;

coef2 = mdl2.Coefficients.Estimate;

fprintf('\nb/x0 from first model: %10.7f\n',coef1(2)/coef1(1))

fprintf('theta from second model: %10.7f\n',coef2(2))

You are seeing, though, that because of the numerical nature of the calculation, they do not return exactly the same result. But I don't think you can "justify" one over the other. They are truly equivalent, mathematically.

It's difficult to tell from inspection, but I think that the second method may be a bit more numerically stable, because with just one parameter in the exponential, you will likely not get as large deviations calculated. Perhaps that is a reason to choose it.

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