curve fit gaussian CDF

I have 20 data points that follow gaussian CDF and would like to find its mean and standard deviation.

If I understand it correctly, normfit is for gaussian PDF.

So, basically, I am looking for a command like normfit but for gaussian CDF.

I did it in Python but could not find a way to do this in matlab.

The picture below is the result from Python.

y = [ 0.0010 0 0.0020 0.0060 0.0210 0.0400 0.0840 0.1890 0.2790 0.4500 0.6180 0.7550 0.8790 0.9330 0.9770 0.9940 0.9980 1.0000 1.0000 1.0000]

x = [0.3100 0.3115 0.3130 0.3145 0.3160 0.3175 0.3190 0.3205 0.3220 0.3235 0.3250 0.3265 0.3280 0.3295 0.3310 0.3325 0.3340 0.3355 0.3370 0.3385]

And this is the result with curve fit app in matlab.

Answer 1

[Star Strider]

Consider using normcdf with mle or fitnlm.

Try this —

y = [    0.0010         0    0.0020    0.0060    0.0210    0.0400    0.0840    0.1890    0.2790    0.4500    0.6180    0.7550    0.8790    0.9330    0.9770  0.9940    0.9980    1.0000    1.0000    1.0000];
x = [0.3100    0.3115    0.3130    0.3145    0.3160    0.3175    0.3190    0.3205    0.3220    0.3235    0.3250    0.3265    0.3280    0.3295    0.3310  0.3325    0.3340    0.3355    0.3370    0.3385];

xv = linspace(min(x), max(x));

ncdf = @(b,x) normcdf(x,b(1),b(2));
ncdfmdl = fitnlm(x,y,ncdf,[0.32 0.01])

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

Estimated Coefficients:
          Estimate         SE        tStat       pValue  
          _________    __________    ______    __________

    b1      0.32394    3.0203e-05     10725    1.0431e-62
    b2    0.0036247    4.2714e-05    84.859    6.9165e-25


Number of observations: 20, Error degrees of freedom: 18
Root Mean Squared Error: 0.00688
R-Squared: 1,  Adjusted R-Squared 1
F-statistic vs. zero model: 9.36e+04, p-value = 7.04e-37

B = ncdfmdl.Coefficients.Estimate;

figure
plot(x, y, 'p')
hold on
plot(xv, ncdf(B,xv), '-r')
hold off
grid

.

[Budi Mulyanto]

Thank you! It works. Is there any idea why it is more straight forward in Python?

[Star Strider]

As always, my pleasure!

‘Is there any idea why it is more straight forward in Python?’

I have no experience with Python, so I can only venture a guess. It may be that Python is relatively young and evolved in a different environment, while MATLAB has been around for several decades (my first experience with it was in 1992 and it was not new then) and so it needs to be backward-compatible with legacy code.

.

Answer 2

[Torsten]

ydata = [    0.0010         0    0.0020    0.0060    0.0210    0.0400    0.0840    0.1890    0.2790    0.4500    0.6180    0.7550    0.8790    0.9330    0.9770  0.9940    0.9980    1.0000    1.0000    1.0000];
xdata = [0.3100    0.3115    0.3130    0.3145    0.3160    0.3175    0.3190    0.3205    0.3220    0.3235    0.3250    0.3265    0.3280    0.3295    0.3310  0.3325    0.3340    0.3355    0.3370    0.3385];
fun = @(mu,sigma) 0.5*(1+erf((xdata-mu)/(sqrt(2)*sigma)));
fun1 = @(mu,sigma) fun(mu,sigma) - ydata;
sol = lsqnonlin(@(x)fun1(x(1),x(2)),[0.325,0.005]);

Local minimum possible.

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

mu = sol(1)

mu = 0.3239

sigma = sol(2)

sigma = 0.0036

plot(xdata,ydata)
hold on
plot(xdata,fun(mu,sigma))

[Budi Mulyanto]

Thank you! your solution also works. But the other answer came slightly earlier. Thank you once again.

[Torsten]

No problem. And although the results are equal, Star Strider's solution has a strong statistical foundation.