以给定的平均值截断法线

问题描述

在python中是否可以生成具有给定期望值的截断正态分布？我知道scipy.stats.truncnorm可以给出以原始正态分布的均值为参数的截断正态分布，但是我想创建一个截断正态分布，以使截短的分布是一个特定值。

Is it possible in python to generate a truncated normal distribution with a given expected value? I know that scipy.stats.truncnorm can give a truncated normal distribution that takes the mean of the original normal distribution as a parameter, but I want to create a truncated normal distribution such that the expected value of the truncated distribution is a particular value. Is this possible?

推荐答案

您可以在 mu 和均值之间进行转换，有关详细信息，请参见，
对于平均值，有一个简单的表达式，要得到 mu ，您必须求解非线性方程

You could convert between mu and mean, see https://en.wikipedia.org/wiki/Truncated_normal_distribution for details,for mean there is simple expression, to get mu you have to solve nonlinear equation

import scipy
from scipy.stats import norm

def get_mean(mu, sigma, a, b):
    alpha = (a - mu)/sigma
    beta  = (b - mu)/sigma
    Z     = norm.cdf(beta) - norm.cdf(alpha)
    return mu + (norm.pdf(alpha) - norm.pdf(beta)) / Z

def f(x, mean, sigma, a, b):
    return mean - get_mean(x, sigma, a, b)

def get_mu(mean, sigma, a, b):
    mu = scipy.optimize.brentq(f, a, b, args=(mean, sigma, a, b))
    return mu

a  = -2.0
b  = 3.0
sigma = 1.0
mu    = 0.0

mean = get_mean(mu, sigma, a, b)
print(mean)

mu = get_mu(mean, sigma, a, b)
print(mu)

从期望的均值得到 mu 后，可以将其放入采样例程

After getting mu from desired mean, you could put it into sampling routine

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