How to use non-linear regression with two input variables?

I have three types of data and all three are matrices of same size. One of them can be expressed as a function of the other two as, Z = aX/ (1 + b * e^(-Y/c)), where X, Y and Z are matrices of same size, say nxm, and I have to find coefficient matrix with a,b & c which would fit the equation. For finding coefficients and fitting them to my data, the code I have written is as follows:

    myfun = @(p,Y,X)  p(1).*X./(1+p(2).*exp(-Y./p(3)));
    param = [3000 0.2 10];

After this I am confused about fitting my data.

Thank you in advance.

Answer 1

[Star Strider]

You have to engage in a bit of deception to fit two independent variables, because the MATLAB curve fitting functions only allow one independent variable in the argument list.

Combine your ‘X’ and ‘Y’ matrices into one variable:

XY(:,:,1) = X;
XY(:,:,2) = Y;

then change ‘myfun’ to:

myfun = @(p,XY)  p(1).*XY(:,:,1)./(1+p(2).*exp(-XY(:,:,2)./p(3)));

I tested this with matrices and it successfully returns a matrix result. I didn’t use it in an actual regression.

[Hema]

Thank you very much for your reply. I tried fitting it as you suggested as follows:

XY(:,:,1) = X;
XY(:,:,2) = PreSx;
myfun = @(p,XY) (p(1).*XY(:,:,1))./(1+p(2).*exp(-XY(:,:,2)./p(3)));
param = [3000 0.2 10];
Coeff = lsqcurvefit(myfun,param,XY,VpDRY);

But after running the script, I got the following error: Local minimum possible.

lsqcurvefit stopped because the final change in the sum of squares relative to 
its initial value is less than the default value of the function tolerance.

<stopping criteria details>

Optimization stopped because the relative sum of squares (r) is changing
by less than options.TolFun = 1.000000e-06.

Optimization Metric                                  Options
relative change r =   2.37e-07              TolFun =   1e-06 (default)

I am not getting what does it mean exactly.

Thank you.

[Star Strider]

My pleasure!

The output does not actually report an error.

It means that the lsqcurvefit found an appropriate parameter estimate that minimised the cost function, but that it thought that the value returned by the cost function was such that you may have enountered a local minimum rather than the global minimum. (The cost function minimisation routines search for a region in which the cost function does not vary much from iteration to iteration. There could be several areas where this occurs, but only one — the global minimum — in which the value of the cost function itself is lower than at any other region.) If the final parameter estimates are what you expect them to be, then the estimate may be correct. If they are not, keep trying different initial parameter estimates until you get the lowest error. To determine this, ask for the second output, ‘resnorm’ in the documentation. That is the norm of the residuals (errors). The lower this is, the more likely the parameter estimates are the best available and that you have found and converged on the global minimum. If it produces nearly the same parameter estimates and ‘resnorm’ values for several different initial parameter estimates, it means that you have found the global minimum, but that the error is higher than expected. The only practical problem with that is that with a relatively high residual error, one or more of your parameter estimates may not be statistically different from zero.

[Hema]

Hi Star Strider,

First of all thanks for such a nice explanation and sorry for the late response. I am sort of new to statistics and your explanation was very helpful for me in understanding the problem in using lsqcurvefit.

Thank you, Hema

[Star Strider]

My pleasure!