如何计算python中正态累积分布函数的倒数？

How do I calculate the inverse of the cumulative distribution function (CDF) of the normal distribution in Python?

如何计算Python中正态分布的累积分布函数（CDF）的倒数？

Which library should I use? Possibly scipy?

我应该使用哪个库？可能是scipy？

2 个解决方案

#1

NORMSINV (mentioned in a comment) is the inverse of the CDF of the standard normal distribution. Using scipy, you can compute this with the ppf method of the scipy.stats.norm object. The acronym ppf stands for percent point function, which is another name for the quantile function.

NORMSINV（在评论中提到）是标准正态分布的CDF的倒数。使用scipy，您可以使用scipy.stats.norm对象的ppf方法计算它。首字母缩写词ppf代表百分点函数，它是分位数函数的另一个名称。

In [20]: from scipy.stats import norm

In [21]: norm.ppf(0.95)
Out[21]: 1.6448536269514722

Check that it is the inverse of the CDF:

检查它是否与CDF相反：

In [34]: norm.cdf(norm.ppf(0.95))
Out[34]: 0.94999999999999996

By default, norm.ppf uses mean=0 and stddev=1, which is the "standard" normal distribution. You can use a different mean and standard deviation by specifying the loc and scale arguments, respectively.

默认情况下，norm.ppf使用mean = 0和stddev = 1，这是“标准”正态分布。您可以分别通过指定loc和scale参数来使用不同的均值和标准差。

In [35]: norm.ppf(0.95, loc=10, scale=2)
Out[35]: 13.289707253902945

If you look at the source code for scipy.stats.norm, you'll find that the ppf method ultimately calls scipy.special.ndtri. So to compute the inverse of the CDF of the standard normal distribution, you could use that function directly:

如果你查看scipy.stats.norm的源代码，你会发现ppf方法最终调用scipy.special.ndtri。因此，要计算标准正态分布的CDF的倒数，您可以直接使用该函数：

In [43]: from scipy.special import ndtri

In [44]: ndtri(0.95)
Out[44]: 1.6448536269514722

#2

# given random variable X (house price) with population muy = 60, sigma = 40
import scipy as sc
import scipy.stats as sct
sc.version.full_version # 0.15.1

#a. Find P(X<50)
sct.norm.cdf(x=50,loc=60,scale=40) # 0.4012936743170763

#b. Find P(X>=50)
sct.norm.sf(x=50,loc=60,scale=40) # 0.5987063256829237

#c. Find P(60<=X<=80)
sct.norm.cdf(x=80,loc=60,scale=40) - sct.norm.cdf(x=60,loc=60,scale=40)

#d. how much top most 5% expensive house cost at least? or find x where P(X>=x) = 0.05
sct.norm.isf(q=0.05,loc=60,scale=40)

#e. how much top most 5% cheapest house cost at least? or find x where P(X<=x) = 0.05
sct.norm.ppf(q=0.05,loc=60,scale=40)

#1

In [20]: from scipy.stats import norm

In [21]: norm.ppf(0.95)
Out[21]: 1.6448536269514722