Why doesn't my Rician in Sympy match the Rician in Scipy?

问题

I've tried to create a Sympy continuous random variable with a Rician distribution. With thanks to help from an earlier question, it seems that the best approach is to subclass SingleContinuousDistribution. I've implemented a distribution that appears to be in agreement between Wikipedia and Scipy, however I am not getting the same results as Scipy.

What follows is code that implements the random variable, extracts its symbolic distribution and converts it to a Numpy representation through lambdify then plots my distribution against the PDF of the Scipy rician distribution.

from sympy import *
from sympy import stats
from scipy import stats as scst
import numpy as np
import matplotlib.pyplot as plt

from sympy.stats.crv_types import rv
from sympy.stats.crv import SingleContinuousDistribution

class RicianDistribution(SingleContinuousDistribution):
    _argnames=('nu','sigma')
    @property
    def set(self): return Interval(0,oo)

    def pdf(self,x):
        nu,sigma=self.nu, self.sigma
        return (x/sigma**2)*exp(-(x**2+nu**2)/(2*sigma**2))*besseli(0,x*nu/sigma**2)

def Rician(name,nu,sigma):
    return rv(name,RicianDistribution,(nu,sigma))

#this line helps lambdify convert the sympy Bessel to a numpy Bessel
printing.lambdarepr.LambdaPrinter._print_besseli=(lambda self,expr: 'i0(%s)'%expr.argument)

x=Symbol('x')  #parameter for density function

sigma=3; pr=4

#create the symbolic Rician and numeric Rician
SpN=Rician('R',pr,sigma)  #signal plus noise
Rsci=scst.rice(pr,scale=sigma)

fx=lambdify(x,stats.density(SpN)(x),'numpy')

xs=np.linspace(0,25,1000)
plt.plot(xs,fx(xs),'b');
plt.plot(xs,Rsci.pdf(xs),'r');

I would expect the results to match, but they don't appear to:

Am I doing something wrong here?

回答1:

The implementation of the Rice distribution in scipy.stats.rice uses a slightly different parameterization than the parameterization described in the wikipedia article.

To make your plots agree, change this line

Rsci=scst.rice(pr,scale=sigma)

Rsci=scst.rice(pr/sigma, scale=sigma)

Here's a longer explanation:

The PDF shown on wikipedia is

The PDF in the docstring of scipy.stats.rice is

However, that formula does not show the scale parameter that all of the continuous distributions in scipy have. (It also doesn't show the location parameter loc, but I'll assume there is no interest in using a Rice distribution with a nonzero location.) To create the formula that includes the scale parameter, we use the standard scale family of a PDF:

So the scipy PDF is actually

If we make the identifications

we obtain the PDF shown in the wikipedia article.

In your code, your parameter pr is ν, so to convert to scipy's parameterization, you must use b = pr / sigma.

来源：https://stackoverflow.com/questions/33049370/why-doesnt-my-rician-in-sympy-match-the-rician-in-scipy

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python