scipy.stats.chi2.ppf() without scipy

Post author By admin
Post date May 4, 2016
4 Comments on scipy.stats.chi2.ppf() without scipy

Allantools has a dependence on scipy because it uses scipy.stats.chi2.ppf() (inverse of the chi-squared cumulative distribution function) in the code for confidence intervals.

When testing scipy takes quite a long time to install and the whole package seems a bit overkill for just this one function.

So I tried implementing it using just numpy. It sort of works, but not for corner cases where p is very close to 1.0 or k is a big number. I get for example:

    top = pow(x, s)*math.exp(-x)*pow(x,k)
OverflowError: (34, 'Numerical result out of range')

1 2	top = pow(x, s)math.exp(-x)pow(x,k) OverflowError: (34, 'Numerical result out of range')

Maybe there's something trick when numerically evaluating lower_gamma() using the series expansion that I am missing...

import scipy.stats
import math
import numpy
	
def bisection(function, k, p, a, b, tol):
	# http://code.activestate.com/recipes/578417-bisection-method-in-python/
	assert (function(a,k)-p)*(function(b,k)-p) < 0 # a, b must bracket root
	c = (a+b)/2.0
	while (b-a)/2.0 > tol:
		if (function(c, k)-p) == 0:
			return c
		elif (function(a,k)-p)*(function(c,k)-p) < 0:
			b = c
		else :
			a = c
		c = (a+b)/2.0
	return c

def lower_gamma(s,x):
	# lower incomplete gamma function
	# https://en.wikipedia.org/wiki/Incomplete_gamma_function#Evaluation_formulae
	g  = 0
	last_g = 1.0
	done = False
	tol = 1.0e-6
	k=0
	while not done:
		top = pow(x, s)*math.exp(-x)*pow(x,k)
		bot = numpy.prod( [float(s+j) for j in range(k+1) ] )
		dg = float(top)/float(bot)
		if dg == float("Inf"):
			break
		g += dg
		k += 1
		if k>100: # get at least 100 terms in the sum
			if g==0:
				break
			delta = abs(dg/g)
			if delta == float("Inf"):
				break
			if delta < tol:
				done = True
		last_g = g
	return g

def chi2_cdf(x, k):
	# chi-squared cumulative density function
	# cdf(x; k) = lower_gamma(k/2, x/2) / gamma(k/2)
	return lower_gamma(k/2.0, x/2.0) / math.gamma(k/2.0)

def chi2_ppf(p, k):
	# chi-squared Percent point function (inverse of cdf percentiles).
	# look for x such that
	# p = chi2_cdf( x=chi2_ppf(p, k), k)
	tol = 1e-8
	lolim = 0
	hilim = k
	while (chi2_cdf(lolim,k)-p)*(chi2_cdf(hilim,k)-p) > 0: 
		hilim *= 1.5
	return bisection( chi2_cdf, k, p, lolim, hilim, tol)

print "scipy cdf: ",scipy.stats.chi2.cdf(55, 33)	
print "own   cdf: ",chi2_cdf(55, 33)


print "scipy ppf ", scipy.stats.chi2.ppf(0.4, 33)
print "  own ppf ", chi2_ppf(0.4, 33)

# test that we really found the inverse
print scipy.stats.chi2.cdf(scipy.stats.chi2.ppf(0.4, 33), 33)
print chi2_cdf( chi2_ppf(0.4, 33), 33 )

# try to check the scipy function against our own function
# for some random input of (p, k)
for n in range(100):
	k = numpy.random.randint(20, 200)
	p = numpy.random.random()
	print k, p,
	a=scipy.stats.chi2.ppf(p, k)
	b=chi2_ppf(p, k)
	ok = numpy.isclose(a, b)
	if ok:
		print  ok
	else:
		print  ok, a, b
	assert ok

import scipy.stats

import math

import numpy

def bisection(function, k, p, a, b, tol):

# http://code.activestate.com/recipes/578417-bisection-method-in-python/

assert (function(a,k)-p)*(function(b,k)-p) < 0 # a, b must bracket root

c = (a+b)/2.0

while (b-a)/2.0 > tol:

if (function(c, k)-p) == 0:

return c

elif (function(a,k)-p)*(function(c,k)-p) < 0:

b = c

else :

a = c

c = (a+b)/2.0

return c

def lower_gamma(s,x):

# lower incomplete gamma function

# https://en.wikipedia.org/wiki/Incomplete_gamma_function#Evaluation_formulae

g = 0

last_g = 1.0

done = False

tol = 1.0e-6

k=0

while not done:

top = pow(x, s)*math.exp(-x)*pow(x,k)

bot = numpy.prod( [float(s+j) for j in range(k+1) ] )

dg = float(top)/float(bot)

if dg == float("Inf"):

break

g += dg

k += 1

if k>100: # get at least 100 terms in the sum

if g==0:

break

delta = abs(dg/g)

if delta == float("Inf"):

break

if delta < tol:

done = True

last_g = g

return g

def chi2_cdf(x, k):

# chi-squared cumulative density function

# cdf(x; k) = lower_gamma(k/2, x/2) / gamma(k/2)

return lower_gamma(k/2.0, x/2.0) / math.gamma(k/2.0)

def chi2_ppf(p, k):

# chi-squared Percent point function (inverse of cdf percentiles).

# look for x such that

# p = chi2_cdf( x=chi2_ppf(p, k), k)

tol = 1e-8

lolim = 0

hilim = k

while (chi2_cdf(lolim,k)-p)*(chi2_cdf(hilim,k)-p) > 0:

hilim *= 1.5

return bisection( chi2_cdf, k, p, lolim, hilim, tol)

print "scipy cdf: ",scipy.stats.chi2.cdf(55, 33)

print "own cdf: ",chi2_cdf(55, 33)

print "scipy ppf ", scipy.stats.chi2.ppf(0.4, 33)

print " own ppf ", chi2_ppf(0.4, 33)

# test that we really found the inverse

print scipy.stats.chi2.cdf(scipy.stats.chi2.ppf(0.4, 33), 33)

print chi2_cdf( chi2_ppf(0.4, 33), 33 )

# try to check the scipy function against our own function

# for some random input of (p, k)

for n in range(100):

k = numpy.random.randint(20, 200)

p = numpy.random.random()

print k, p,

a=scipy.stats.chi2.ppf(p, k)

b=chi2_ppf(p, k)

ok = numpy.isclose(a, b)

if ok:

print ok

else:

print ok, a, b

assert ok

Tags allantools, python, scipy

4 replies on “scipy.stats.chi2.ppf() without scipy”

Try computing dg using log-math. not sure if this will render in a comment or not:

        while not done:
                if x != 0:
                    log_top = math.log(x) * (s + k) - x
                else:
                    log_top = 0
                log_bot = sum(math.log(s+j) for j in range(k+1))
                dg = math.exp(log_top - log_bot)

while not done:

if x != 0:

log_top = math.log(x) * (s + k) - x

else:

log_top = 0

log_bot = sum(math.log(s+j) for j in range(k+1))

dg = math.exp(log_top - log_bot)

however, this has some failures of the 'isclose' test e.g., at
190 0.9999 False 271.179788314 271.221549805
(relative error .00015)

full test program adapted from yours at https://emergent.unpythonic.net/files/sandbox/chi2_ppf_jepler.py

boost discusses their implementation: http://www.boost.org/doc/libs/1_35_0/libs/math/doc/sf_and_dist/html/math_toolkit/special/sf_gamma/igamma.html#math_toolkit.special.sf_gamma.igamma.implementation using several different methods depending on the values of (what they call) x and a, may be worth a gander.

Thanks! I will take a look at the links.

HTML tags 'pre lang="python"' around the code makes it render using the WP-Syntax plugin.

You could maybe get away with exploiting the fact that the gamma function in the denominator of the pdf of a chi-squared random variable is just a normalizing constant. So, e.g., these three functions get me pretty close to the output of scipy.stats.chi2.ppf():

def chi2_pdf(x,k):
    bin = x[1]-x[0]
    fx = (x**((k/2)-1)*np.exp(-x/2))/(2**(k/2))
    fx = fx/(bin*np.sum(fx))
    return fx, bin

def chi2_cdf(x,k):
    pdf,bin = chi2_pdf(x,k)
    cdf = np.zeros(len(x))
    for i in range(len(x)):
        cdf[i] = bin*np.sum(pdf[:i])
    return cdf

def chi2_inv(p,x,k):
    cdf = chi2_cdf(x,k)
    ix = np.argmin(np.abs(cdf-p))
    return x[ix]

def chi2_pdf(x,k):

bin = x[1]-x[0]

fx = (x**((k/2)-1)*np.exp(-x/2))/(2**(k/2))

fx = fx/(bin*np.sum(fx))

return fx, bin

def chi2_cdf(x,k):

pdf,bin = chi2_pdf(x,k)

cdf = np.zeros(len(x))

for i in range(len(x)):

cdf[i] = bin*np.sum(pdf[:i])

return cdf

def chi2_inv(p,x,k):

cdf = chi2_cdf(x,k)

ix = np.argmin(np.abs(cdf-p))

return x[ix]

The "binsize" of x matters, both for speed and accuracy. And if you don't know what k is ahead of time, the fact that you have to define x is kind of a pain, though you can probably also exploit the fact that the expected value of a chi-squared random variable is k and the variance is 2k to define x over an appropriate range once k is known.

Here's hoping I used the "pre" tag correctly...

I have a feeling that the cdf(x,k) function is very flat at high x as we get closer to p=1.0. Conversely ppf(p,k) is very steep as we get closer to p=1.0

One improvement could be to change the bisection() function that looks for the right value of x to instead stop searching with a tolerance check on p (instead of x). Maybe?

4 replies on “scipy.stats.chi2.ppf() without scipy”

Leave a Reply Cancel reply