用于约束参数的卡方检验 [英] Chi-square testing for constraining a parameter
问题描述
我有一个关于使用 chi^2 检验来约束宇宙学参数的重要问题.我感谢您的帮助.请不要给这个问题负分(这个问题对我很重要).
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).
假设我们有一个包含 600 个数据的数据文件 (data.txt)，该数据文件有 3 列，第一列是 redshift(z)，第二列是观察 dL(m_obs)，第三列是列是错误(错误).我们知道 chi^2 函数是
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
我们必须计算的所有事情是将给定数据文件中的 z 放入我们在 m_theo 中的函数中，用于所有 600 个数据并计算 chi^2.现在在 m_thoe 中我们有一个自由参数(o_m)，我们必须找到它的值，其中 chi^2 达到其最小值.
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
这个问题不是重复的，对每个使用 chi^2 的身体都非常重要，特别是宇宙学家和物理学家.如何找到最小化的 chi^2 和相对 o_m?
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)
我认为我的代码是正确的，但它没有给我正确的答案.谢谢您，感谢您的时间和关注.
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.
推荐答案
这是正确答案:
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)
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