问题
I have an important question about the use of chi^2 test to constrain a parameter in cosmology. I appreciate your help. Please do not give this question negative rate (this question is important to me).
Assume we have a data file (data.txt
) concluding 600 data and this data file has 3 columns, first column is redshift(z), second column is observational dL(m_obs) and third column is error(err). As we know chi^2 function is
chi^2=(m_obs-m_theo)**2/err**2 #chi^2=sigma((m_obs-m_theo)**2/err**2) from 1 to N=600
All of thing that we must calculate is putting z
from given data file into our function in m_theo
for all 600 data and calculate chi^2. Now in m_thoe
we have a free parameter(o_m
) and we have to find its value in which the chi^2 reach its minimum value.
q= 1/sqrt( (1+z)**2 * (1+0.01*o_m*z) - z*(2+z)*(1-0.01*o_m) )
m_theo = 5.0 * log10( (1+z)*q ) + 43.1601
This question is not repetitive and is very important to every body using chi^2 specially for cosmologists and physicists.
How to find minimized chi^2 and relative o_m
?
from math import *
import numpy as np
from scipy.integrate import quad
min=l=a=b=chi=None
c=0 #for Sigma or summation chi^2 terms in c=c+chi for first term
def ant(z,o_m): #0.01*o_m is steps of o_m
return 1/sqrt(((1+z)**2*(1+0.01*o_m*z)-z*(2+z)*(1-0.01*o_m)))
for o_m in range(24,35,1): #arbitrary range of o_m
############## opening data file containing 580 dataset
with open('data.txt') as f:
for i, line in enumerate(f): #
n= list(map(float, line.split())) #
for i in range(1):
##############
q=quad(ant,0,n[1],args=(o_m,))[0] #Integration o to z, z=n[1]
h=5*log10((1+n[1])*(299/70)*q)+25 #function of dL
chi=(n[2]-h)**2/n[3]**2 #chi^2 test function
c=c+chi #sigma from 1 to N of chi^2 and N=580
if min is None or min>c:
min=c
print(c,o_m)
I think my code is correct but it doesn't give me a proper answer. Thank you and I appreciate your time and your attention.
回答1:
This is the correct answer:
from math import *
import numpy as np
from scipy.integrate import quad
min=l=a=b=chi=None
c=0
z,mo,err=np.genfromtxt('Union2.1_z_dm_err.txt',unpack=True)
def ant(z,o_m): #0.01*o_m is steps of o_m
return 1/sqrt(((1+z)**2*(1+0.01*o_m*z)-z*(2+z)*(1-0.01*o_m)))
for o_m in range(20,40):
c=0
for i in range(len(z)):
q=quad(ant,0,z[i],args=(o_m,))[0] #Integration o to z
h=5*log10((1+z[i])*(299000/70)*q)+25 #function of dL
chi=(mo[i]-h)**2/err[i]**2 #chi^2 test function
c=c+chi
l=o_m
print('chi^2=',c,'Om=',0.01*l,'OD=',1-0.01*l)
来源:https://stackoverflow.com/questions/45631732/chi-square-testing-for-constraining-a-parameter