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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

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the cyclist
the cyclist 2021-5-5

1 个投票

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.

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Collin Pecora
Collin Pecora 2021-5-5
Cyclist
Your insight is absolutely correct, both that the models are mathematically equivalent and for what the x0 parameter does.
Let’s say you are using a minimizer, like say, fmincon where the objective function is the sum of squared errors (SSE) between your measurements and either model1 or model2. Note that x0 is one of the key parameters we are after. Both models will result in nearly identical x0's with model1 usually terminating at slightly smaller SSEs, thus indicating that model1 has a small advantage over model2 using SSE as a scoring metric.
My question is why, what is it about wrapping the x0 into the theta parameter in model2 that causes the slight differences in results, and is there a way I can express it numerically or statistically?
the cyclist
the cyclist 2021-5-6
Ran in:
Ran in:
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])
mdl1 =
Nonlinear regression model: y ~ (1 - (x/B11))*exp( - B12*(x/B11)) Estimated Coefficients: Estimate SE tStat pValue ________ ______ _______ _______ B11 3.2994 2.388 1.3817 0.18399 B12 2.1569 2.4787 0.87016 0.39567 Number of observations: 20, Error degrees of freedom: 18 Root Mean Squared Error: 0.0298 R-Squared: 0.979, Adjusted R-Squared 0.978 F-statistic vs. zero model: 4.99e+03, p-value = 1.98e-25
mdl2 = fitnlm(x,y,f2,[1 1])
mdl2 =
Nonlinear regression model: y ~ (1 - (x/B21))*exp( - B22*x) Estimated Coefficients: Estimate SE tStat pValue ________ _______ ______ ________ B21 3.2994 2.3888 1.3812 0.18412 B22 0.65372 0.27883 2.3445 0.030724 Number of observations: 20, Error degrees of freedom: 18 Root Mean Squared Error: 0.0298 R-Squared: 0.979, Adjusted R-Squared 0.978 F-statistic vs. zero model: 4.99e+03, p-value = 1.98e-25
% 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))
b/x0 from first model: 0.6537185
fprintf('theta from second model: %10.7f\n',coef2(2))
theta from second model: 0.6537189
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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