Fitting data with a exponentially modified gaussian equation
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Hello,
I am using curve fitting tool box and I am trying to fit below equation of the exponentially modified gaussian distribution. I have attached my attempt at fittin the PDF, I have also attached my x and y data along with this question. Please help me with the solution. Thanks very much!
(c/2) * (exp((c^2*sg^2)/2)) * exp(-c*(x-mu)) * erf((1/sqrt(2)) * ((c*sg) - ((x-mu)/sg)))
回答（1 个）
Because you are using a custom model, you need to supply an initial guess of the parameters under "Fit Options". Otherwise, the toolbox will choose one at random (which will probably be bad).
5 个评论
Sanchit Sharma
2022-3-18
编辑：Sanchit Sharma
2022-3-18
Torsten
2022-3-18
Plot the function for values of mu, c and sb and compare with your data curve.
This will give you a feeling about the form of the curve and a good starting guess.
Are you sure that the integral from -oo to oo will be 1 for all choices of mu, c and sg as is necessary for a pdf ?
Matt J
2022-3-18
Plot the function for values of mu, c and sb and compare with your data curve.
Or, compute the fitting error for different combinations of mu,c, sg. You can do a grid search on, say, a 15x15x15 grid of mu,c,sg, values. Once you find the best fit from the grid search, you can use that as your starting guess in a more formal fit.
Torsten
2022-3-18
Is your function a probability density function ?
If yes, try
omega = 800.0;
zeta = 7500;
alpha = -1.3;
f = @(x) 2/(omega*sqrt(2*pi))*exp(-((x-zeta).^2/(2*omega^2)))*0.5.*(1+erf(alpha*(x-zeta)/omega));
x = 7000-1500:0.01:7000+1000;
plot(x,f(x))
But your .png file seems to indicate that the area under your curve is not equal to 1. So I assume you are not dealing with a pdf. In this case, the above function f is not suitable.
John D'Errico
2022-3-18
It vaguely looks like the data comes from a histogram, but it was not normalized to have unit area under the curve.
trapz(x,y)
ans =
73.2481
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