Reputation: 4318
I want to fit an asymmetric probability distribution to my data and I thought an exponentially modified Gaussian distribution can be a good representative for my data. I
m=array([ 16.25, 16.75, 17.25, 17.75, 18.25, 18.75, 19.25, 19.75,
20.25, 20.75, 21.25, 21.75, 22.25, 22.75, 23.25, 23.75,
24.25, 24.75, 25.25, 25.75, 26.25, 26.75, 27.25, 27.75,
28.25, 28.75, 29.25, 29.75, 30.25, 30.75])
pdf=array([ 0.00000000e+00, 2.40818784e-04, 1.38470801e-03,
1.62552679e-03, 3.07043949e-03, 3.37146297e-03,
5.47862733e-03, 8.36845274e-03, 1.61348585e-02,
1.92052980e-02, 2.79951836e-02, 3.97953040e-02,
4.95484648e-02, 7.09211318e-02, 9.50030102e-02,
1.40878989e-01, 1.90186635e-01, 2.42022878e-01,
2.77302830e-01, 2.69054786e-01, 2.40397351e-01,
1.74593618e-01, 9.16917520e-02, 2.41420831e-02,
7.22456352e-03, 3.01023480e-04, 0.00000000e+00,
0.00000000e+00, 0.00000000e+00, 6.02046960e-05])
I would like to use scipy.optimize library and in meanwhile could control the goodness of fit maybe and see it in order to improve the chi-squared by changing the initial conditions for input parameters. I have written the following piece of code:
import scipy.special as sse
from math import *
import numpy as np
import scipy.optimize
#defines the PDF of an exponentially modified Gaussian distribution
fitfunc =lambda p,x: 0.5*p[2]*np.exp(0.5*p[2]*(2*p[0]+p[2]*p[1]*p[1]-2*x))*sse.erfc((p[0]+p[2]*p[1]*p[1]-x)/(np.sqrt(2)*p[1]))
"""Deviations of data from fitted curve"""
errfunc = lambda p, x, y: fitfunc(p, x) - y
#initial values
p0=[24,1,1]
p1, success = scipy.optimize.leastsq(errfunc, p0, args=(pdf, m), maxfev=10000)
Update: well I just chose numpy.exp
and the first problem solved but still leastsq doesn't give me reliable outputs, what should I do? In addition, I would like to obtain the CDF for this distribution too.
Upvotes: 1
Views: 921
Reputation: 17585
The method of least squares IS NOT the method to use for fitting data to a given pdf.
What you (probably) want is the method of maximum likelihood -- i.e. to maximize p(x | a) where a are the parameters of the distribution and x are the data. Typically one forms the log-likelihood and assumes independence, so log p(x | a) = sum(log(pdf(x[i], a)), i, 1, n).
You need to use a minimization function, giving it log p(x | a) as the function to be minimized, and a as its free parameters.
Upvotes: 4