[Numpy-discussion] Power distribution
josef.pktd at gmail.com
josef.pktd at gmail.com
Fri Aug 7 20:54:49 EDT 2009
On Fri, Aug 7, 2009 at 6:57 PM, <josef.pktd at gmail.com> wrote:
> On Fri, Aug 7, 2009 at 6:13 PM, <josef.pktd at gmail.com> wrote:
>> On Fri, Aug 7, 2009 at 5:42 PM, <josef.pktd at gmail.com> wrote:
>>> On Fri, Aug 7, 2009 at 5:25 PM, Andrew Hawryluk<HAWRYLA at novachem.com> wrote:
>>>> Hmm ... good point.
>>>> It appears to give a probability distribution proportional to x**(a-1),
>>>> but I see no good reason why the domain should be limited to [0,1].
>>>>
>>>> def test(a):
>>>> nums =
>>>> plt.hist(np.random.power(a,100000),bins=100,ec='none',fc='#dddddd')
>>>> x = np.linspace(0,1,200)
>>>> plt.plot(x,nums[0][-1]*x**(a-1))
>>>>
>>>> Andrew
>>>>
>>>>
>>>>
>>>>> -----Original Message-----
>>>>> From: numpy-discussion-bounces at scipy.org [mailto:numpy-discussion-
>>>>> bounces at scipy.org] On Behalf Of alan at ajackson.org
>>>>> Sent: 7 Aug 2009 2:49 PM
>>>>> To: Discussion of Numerical Python
>>>>> Subject: Re: [Numpy-discussion] Power distribution
>>>>>
>>>>> I don't think that is it, since the one in numpy has a range
>>>> restricted
>>>>> to the interval 0-1.
>>>>>
>>>>> Try out hist(np.random.power(5, 1000000), bins=100)
>>>>>
>>>>> >You might get better results for 'power-law distribution'
>>>>> >http://en.wikipedia.org/wiki/Power_law
>>>>> >
>>>>> >Andrew
>>>>> >
>>>>> >> -----Original Message-----
>>>>> >> From: numpy-discussion-bounces at scipy.org [mailto:numpy-discussion-
>>>>> >> bounces at scipy.org] On Behalf Of alan at ajackson.org
>>>>> >> Sent: 7 Aug 2009 11:45 AM
>>>>> >> To: Discussion of Numerical Python
>>>>> >> Subject: [Numpy-discussion] Power distribution
>>>>> >>
>>>>> >> Documenting my way through the statistics modules in numpy, I ran
>>>>> >> into the Power Distribution.
>>>>> >>
>>>>> >> Anyone know what that is? I Googled for it, and found a lot of
>>>> stuff
>>>>> >on
>>>>> >> electricity, but no reference for a statistical distribution of
>>>> that
>>>>> >> name. Does it have a common alias?
>>>>> >>
>>>>> >> --
>>>
>>>
>>> same is in Travis' notes on the distribution and scipy.stats.distributions
>>> domain in [0,1], but I don't know anything about it either
>>>
>>> ## Power-function distribution
>>> ## Special case of beta dist. with d =1.0
>>>
>>> class powerlaw_gen(rv_continuous):
>>> def _pdf(self, x, a):
>>> return a*x**(a-1.0)
>>> def _cdf(self, x, a):
>>> return x**(a*1.0)
>>> def _ppf(self, q, a):
>>> return pow(q, 1.0/a)
>>> def _stats(self, a):
>>> return a/(a+1.0), a*(a+2.0)/(a+1.0)**2, \
>>> 2*(1.0-a)*sqrt((a+2.0)/(a*(a+3.0))), \
>>> 6*polyval([1,-1,-6,2],a)/(a*(a+3.0)*(a+4))
>>> def _entropy(self, a):
>>> return 1 - 1.0/a - log(a)
>>> powerlaw = powerlaw_gen(a=0.0, b=1.0, name="powerlaw",
>>> longname="A power-function",
>>> shapes="a", extradoc="""
>>>
>>> Power-function distribution
>>>
>>> powerlaw.pdf(x,a) = a*x**(a-1)
>>> for 0 <= x <= 1, a > 0.
>>> """
>>> )
>>>
>>
>>
>> it looks like it's the same distribution, even though it doesn't use
>> the random numbers from the numpy function
>>
>> high p-values with Kolmogorov-Smirnov, see below
>>
>> I assume it is a truncated version of *a* powerlaw distribution, so
>> that a can be large, which would be impossible in the open domain
>> case. But a quick search, I only found powerlaw applications that
>> refer to the tail behavior.
>>
>> Josef
>>
>>>>> rvs = np.random.power(5, 100000)
>>>>> stats.kstest(rvs,'powerlaw',(5,))
>> (0.0021079715221341555, 0.76587118275752697)
>>>>> rvs = np.random.power(5, 1000000)
>>>>> stats.kstest(rvs,'powerlaw',(5,))
>> (0.00063983013407076239, 0.80757958281509501)
>>>>> rvs = np.random.power(0.5, 1000000)
>>>>> stats.kstest(rvs,'powerlaw',(0.5,))
>> (0.00081823148457027539, 0.51478478398950211)
>>
>
> I found a short reference in Johnson, Kotz, Balakrishnan vol. 1 where
> it is refered to as the "power-function" distribution.
> roughly: if X is pareto (which kind) distributed, then Y=X**(-1) is
> distributed according to the power-function distribution. JKB have an
> extra parameter in there and is a bit more general then the scipy
> version, or maybe it is just the scale parameter included in the
> density function.
>
> It is also in NIST data plot, but I didn't find the html reference
> page, but only the pdf
>
> http://docs.google.com/gview?a=v&q=cache%3AEgQ6bRkeJl8J%3Awww.itl.nist.gov%2Fdiv898%2Fsoftware%2Fdataplot%2Frefman2%2Fauxillar%2Fpowpdf.pdf+power-function+distribution&hl=en&gl=ca&pli=1
>
> the pdf-files for powpdf and powcdf are here
> http://www.itl.nist.gov/div898/software/dataplot/refman2/auxillar/homepage.htm
>
>
> I can look some more a bit later tonight.
>
> Josef
>
for the relationship to pareto, below are some kstests and graphs
a reminder that numpy.random.pareto uses a non-standard 0 bound, instead of 1
ks tests don't show good numbers every once in a while, since they are random
I checked the definitions in JKB (page 607) and my previous
interpretation was correct.
if X has a pareto distribution with lower bound at 1 and shape
parameter a>0, then 1/X has a density function
p(y) = a*y**(a-1), (0<y<1)
weak inequality in JKB instead of strict as in scipy.stats.powerlaw docstring
(the actual scipy.stats.powerlaw docstring has a typo, a**x**(a-1),
which I will correct)
Josef
import numpy as np
from scipy import stats
import matplotlib.pyplot as plt
rvs = np.random.power(5, 1000000)
rvsp = np.random.pareto(5, 1000000)
rvsps = stats.pareto.rvs(5, size=100)
print "stats.kstest(1./rvsps,'powerlaw',(5,))"
print stats.kstest(1./rvsps,'powerlaw',(5,))
print "stats.kstest(1./(1+rvsp),'powerlaw',(5,))"
print stats.kstest(1./(1+rvsp),'powerlaw',(5,))
print "stats.kstest(rvs,'powerlaw',(5,))"
print stats.kstest(rvs,'powerlaw',(5,))
print "stats.ks_2samp(rvs,1./(rvsp+1))"
print stats.ks_2samp(rvs,1./(rvsp+1))
print "stats.ks_2samp(rvs,1./rvsps)"
print stats.ks_2samp(rvs,1./rvsps)
print "stats.ks_2samp(1+rvsp, rvsps)"
print stats.ks_2samp(1+rvsp, rvsps)
plt.figure()
plt.hist(rvs, bins=50)
plt.title('np.random.power(5)')
plt.figure()
plt.hist(1./(1.+rvsp), bins=50)
plt.title('inverse of 1 + np.random.pareto(5)')
plt.figure()
plt.hist(1./(1.+rvsp), bins=50)
plt.title('inverse of stats.pareto(5)')
#plt.show()
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