Fitting a pareto distribution with (python) Scipy
I have a data set that I know has a Pareto distribution. Can someone point me to how to fit this data set in Scipy? I got the below code to run but I have no idea what is being returned to me (a,b,c). Also, after obtaining a,b,c, how do I calculate the variance using them?
import scipy.stats as ss
import scipy as sp
a,b,c=ss.pareto.fit(data)
Here's a quickly written version, taking some hints from the Reference page that Rupert gave. This is currently work in progress in scipy and statsmodels and requires MLE with some fixed or frozen parameters, which is only available in the trunk versions. No standard errors on the parameter estimates or other result statistics are available yet.
'''estimating pareto with 3 parameters (shape, loc, scale) with nested
minimization, MLE inside minimizing Kolmogorov-Smirnov statistic
running some examples looks good
Author: josef-pktd
'''
import numpy as np
from scipy import stats, optimize
#the following adds my frozen fit method to the distributions
#scipy trunk also has a fit method with some parameters fixed.
import scikits.statsmodels.sandbox.stats.distributions_patch
true = (0.5, 10, 1.) # try different values
shape, loc, scale = true
rvs = stats.pareto.rvs(shape, loc=loc, scale=scale, size=1000)
rvsmin = rvs.min() #for starting value to fmin
def pareto_ks(loc, rvs):
est = stats.pareto.fit_fr(rvs, 1., frozen=[np.nan, loc, np.nan])
args = (est[0], loc, est[1])
return stats.kstest(rvs,'pareto',args)[0]
locest = optimize.fmin(pareto_ks, rvsmin*0.7, (rvs,))
est = stats.pareto.fit_fr(rvs, 1., frozen=[np.nan, locest, np.nan])
args = (est[0], locest[0], est[1])
print 'estimate'
print args
print 'kstest'
print stats.kstest(rvs,'pareto',args)
print 'estimation error', args - np.array(true)
Be very careful fitting power laws!! Many reported power laws are actually badly fitted by a power law. See Clauset et al. for all the details (also on arxiv if you don't have access to the journal). They have a companion website to the article which now links to a Python implementation. Don't know if it uses Scipy because I used their R implementation when I last used it.
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