Python Scipy 中的麦克斯韦分布

Question

In the article that I am interested in, it states that the data is well represented with a Maxwellian distribution and it also provides a Mean speed (307 km/s) and 1 sigma uncertainty (47 km/s) for the distribution.

Using the provided values, I have attempted to re-generate the data and then fit it with the Maxwellian distribution using the python scipy.stats.

As it described in here , maxwell function in scipy takes two input, 1) "loc" which shifts the x variable and 2) "a" parameter which corresponds to the parameter "a" in the maxwell-Boltzmann equation.

In my case, I have neither of these parameters, so using the Mean and variance (sigma^2) description in wiki page , I have attempted to calculate the "a" and "loc" parameter. Both mean and sigma parameters are only dependent on "a" parameter.

The first problem I have encountered was the "a" parameter that I get from Mean (a = 192.4) and sigma (a = 69.8) are different from each other. The second problem is that I don't know how can I obtain the exact loc (shift) value from Mean and sigma.

Based on the shape of the distribution (where mean speed values fall in the graph, check figure 2), I tried to guess the "loc" value and together with the "a" value obtained from sigma (a = 69.8), I have generated and fitted the data. Approximately it seems correct, but I don't know the answer to the questions I mentioned above and I need some expert's guidance on this. I appreciate any help.

import matplotlib.pyplot as plt
import math
from scipy.stats import norm
import random
import numpy as np
import scipy.optimize
from scipy.stats import maxwell

samplesize = 100000

mean = 307
sigma = 47
loc = 175 #my guess
a_value = np.sqrt((sigma**2 * math.pi)/(3*math.pi - 8)) #calculated based on wiki description

fig, axs = plt.subplots(1)
v_2d = maxwell.rvs(loc, a_value, size=samplesize) #array corresponding to 2D proper motion obtained from Hubbs
mean, var, skew, kurt = maxwell.stats(moments='mvsk')

N, bins, patches = plt.hist(v_2d, bins=100, density=True, alpha=0.5, histtype='bar', ec='black')
maxx = np.linspace(min(v_2d), max(v_2d), samplesize)

axs.plot(maxx, maxwell.pdf(maxx, loc, a_value), color=colorset[6], lw=2, label= r'$\mathdefault{\mu}$ = '+'{:0.1f}'.format(mean)+r' , '+r'$\mathdefault{\sigma}$ = '+'{:0.1f}'.format(sigma))

axs.set(xlabel=r'2-D Maxwellian speed (km s$^{-1}$)')
axs.set(ylabel='Frequency')
plt.legend(loc='upper right')

Answer 1

Well, mean value is affected by location, and sigma won't. So compute a from sigma, compute mean as if loc=0, find the difference and assign it to location, sample 100K RVs to check if sampled mean/stddev are close enough.

Code, Python 3.8, Windows 10 x64

import numpy as np

from scipy.stats import maxwell

σ = 47
μ = 307

a = σ * np.sqrt(np.pi/(3.0*np.pi - 8.0))
print(a)

m = 2.0*a*np.sqrt(2.0/np.pi)
print(m) # as if loc=0

loc = μ - m
print(loc)

print("----------Now test--------------------")

# sampling
q = maxwell.rvs(loc=loc, scale=a, size=100000)

print(np.mean(q))
print(np.std(q))

as output I've got

306.9022249667151
47.05319429681308

Good enough?