Fitting Cumulative Gaussian Function to data

Hello ,

I am trying to fit the cumulative Gaussian Function to my data points, to find out the PSE. So far I used this function:

f = @(b,x) normcdf(x, b(1), b(2)); % Objective Function

NRCF = @(b) norm(y - f(b,x)); % Norm Residual Cost Function

B1 = fminsearch(NRCF, [-1; 5]); % Estimate Parameters

x = 0:5

y = [0.7 0.4 0.2 0.4 0.6 0.9]

I am getting: 1.91293286

However, the correct value should be: 1.81051. Does anyone have ideas? Is there a better way to fit the cumulative function to my data in Matlab?

Thank you for your help :)

[the cyclist]

After I posted my answer, I happened to notice that you already had a post about this question here. @Matt J also pointed out that the model is a very poor one for your data.

Also, what is "PSE"? I don't recognize that in this context.

Answer 1

[the cyclist]

I get the same coefficient that you did, using the fitnlm function instead.

This seems to be a terrible model for your data, which is obvious when you plot them. (Always plot the data!) Why did you choose it?

f = @(b,x) normcdf(x, b(1), b(2));

x = 0:5;
y = [0.7 0.4 0.2 0.4 0.6 0.9];

mdl = fitnlm(x,y,f,[1 1])

mdl = 
Nonlinear regression model:
    y ~ normcdf(x,b1,b2)

Estimated Coefficients:
          Estimate      SE       tStat     pValue 
          ________    ______    _______    _______

    b1     1.9129      2.238    0.85476    0.44086
    b2     7.8396     9.9659    0.78664    0.47548


Number of observations: 6, Error degrees of freedom: 4
Root Mean Squared Error: 0.259
R-Squared: 0.143,  Adjusted R-Squared -0.0709
F-statistic vs. zero model: 13, p-value = 0.0177

xq = (-2 : 0.1 : 8)';
yq = predict(mdl,xq);

figure
hold on
hd = plot(x,y,'.');
set(hd,'MarkerSize',24)
hf = plot(xq,yq);
set(hf,'LineWidth',2)
ylim([0 1])
legend([hd,hf],{'data','fit'},'Location','SouthWest')

[Sarah T]

Hello,

thank you for your quick answer and your time! That is an interesting model you propose here.

If I see it correctly you firstly used the gaussian function and then introduced another fit parameter? Afterwards you introduced a prediction model? I tried to use this model for my other data, but for some it did not work.

It is true, I asked this question already. However, I had some confusing about the paramters, so the answer was not quite satisfying for me. Hence, I wanted to acknowdlege the work the other person did.

[Sarah T]

e.g. y = 0.100000000000000 0 0 1 1 1

Error using nlinfit>checkFunVals (line 649)

The function you provided as the MODELFUN input has returned Inf or NaN values.

Error in nlinfit>LMfit (line 596)

if funValCheck && ~isfinite(sse), checkFunVals(r); end

Error in nlinfit (line 284)

[beta,J,~,cause,fullr] = LMfit(X,yw, modelw,beta,options,verbose,maxiter);

Error in NonLinearModel/fitter (line 1127)

nlinfit(X,y,F,b0,opts,wtargs{:},errormodelargs{:});

Error in classreg.regr.FitObject/doFit (line 94)

model = fitter(model);

Error in NonLinearModel.fit (line 1446)

model = doFit(model);

Error in fitnlm (line 99)

model = NonLinearModel.fit(X,varargin{:});

[the cyclist]

I'm a little confused by your remark, "That is an interesting model you propose here."

The model is exactly the same as in your original question. I just used a different MATLAB function to find the best fit. There is no "extra" fitting parameter. Your code reports the two coefficients, just the same:

x = 0:5 

x = 1×6

0 1 2 3 4 5

y = [0.7 0.4 0.2 0.4 0.6 0.9];

f = @(b,x) normcdf(x, b(1), b(2)); % Objective Function 
NRCF = @(b) norm(y - f(b,x)); % Norm Residual Cost Function 
B1 = fminsearch(NRCF, [-1; 5]) % Estimate Parameters

B1 = 2×1

1.9129 7.8395

And then, since this is a model that fits a model of the form f(x) = y, all I do in the last part is create a sequence of uniformly space x values, and find the "predicted" y values given by the fitted equation. You could have done that same with your output. Here is a tiny example:

xq = [-1 0 1];
yq = f(B1,xq)

yq = 1×3

0.3551 0.4036 0.4536

I still don't understand why you are trying to fit this function to the data points, which are shaped nothing like a CDF.

[Sarah T]

Ah, its true. Basically you also get the same value with 1.91 and not 1.81.

I have a data set for whom I would like calculate psychometric functions. With 80% of my data this works fine, but for some data points I get different values compared to fitiing with CDF in other programs. This counts especially for data that also do not really look like CDF (like the example I posted). That is why I thought I am not using a good model right now in Matlab and there would be possibilities to change the fit.