Python 为什么statistics.variance默认使用“无偏”样本方差?
我最近开始使用python的统计模块 我注意到,默认情况下,方差方法返回“无偏”方差或样本方差:Python 为什么statistics.variance默认使用“无偏”样本方差?,python,statistics,variance,Python,Statistics,Variance,我最近开始使用python的统计模块 我注意到,默认情况下,方差方法返回“无偏”方差或样本方差: import statistics as st from random import randint def myVariance(data): # finds the variance of a given set of numbers xbar = st.mean(data) return sum([(x - xbar)**2 for x in data])/len(d
import statistics as st
from random import randint
def myVariance(data):
# finds the variance of a given set of numbers
xbar = st.mean(data)
return sum([(x - xbar)**2 for x in data])/len(data)
def myUnbiasedVariance(data):
# finds the 'unbiased' variance of a given set of numbers (divides by N-1)
xbar = st.mean(data)
return sum([(x - xbar)**2 for x in data])/(len(data)-1)
population = [randint(0, 1000) for i in range(0,100)]
print myVariance(population)
print myUnbiasedVariance(population)
print st.variance(population)
输出:
81295.8011
82116.9708081
82116.9708081
我觉得这很奇怪。我猜很多时候人们都在处理样本,所以他们想要样本方差,但我希望使用默认函数来计算总体方差。有人知道这是为什么吗?我认为,几乎所有的时候,当人们从数据中估计方差时,他们都使用样本。根据无偏估计的定义,方差无偏估计的期望值等于总体方差 在您的代码中,使用random.randint0,1000,它从离散均匀分布中采样1001个可能值和方差1000*1002/12=83500参见,例如。这里的代码显示,平均而言,当使用样本作为输入时,statistics.variance比statistics.pvariance更接近总体方差:
import statistics as st
from random import randint
def myVariance(data):
# finds the variance of a given set of numbers
xbar = st.mean(data)
return sum([(x - xbar)**2 for x in data])/len(data)
def myUnbiasedVariance(data):
# finds the 'unbiased' variance of a given set of numbers (divides by N-1)
xbar = st.mean(data)
return sum([(x - xbar)**2 for x in data])/(len(data)-1)
population = [randint(0, 1000) for i in range(0,100)]
print myVariance(population)
print myUnbiasedVariance(population)
print st.variance(population)
这里是示例输出:
mean variance(sample): 83626.0
mean pvariance(sample): 75263.4
pvariance(population): 83500.0
这里是另一个伟大的职位。我想知道完全相同的事情,这个问题的答案真的为我澄清了。使用np.var,您可以向其中添加一个参数ddof=1,以返回无偏估计值。请查看:
我猜是最小惊喜原则
print(np.var([1,2,3,4],ddof=1))
1.66666666667