如何从(任意)连续概率分布中模拟？

Question

我有这样的概率密度函数：

def p1(x):
    return ( sin(x) ** (-0.75) ) / (4.32141 * (x ** (1/5)))

我想用这个[0; 1]来去除pdf上的随机值。我怎么做随机值？

Answer 1

正如方济各提到的，你最好知道你的分布情况。总之，way为定义自定义发行版提供了一种方便的方法。看上去就像那样

from scipy import stats
class your_distribution(stats.rv_continuous):
    def _pdf(self, x):
        return ( sin(x) ** (-0.75) ) / (4.32141 * (x ** (1/5)))

distribution = your_distribution()
distribution.rvs()

Answer 2

不使用can并给出PDF的数值抽样，您可以使用累积分布和线性插值进行抽样。下面的代码在x中假设相同的间距。它可以被修改为对任意采样的PDF进行集成。注意，它在x的范围内将PDF重命名为1。

import numpy as np

def randdist(x, pdf, nvals):
    """Produce nvals random samples from pdf(x), assuming constant spacing in x."""

    # get cumulative distribution from 0 to 1
    cumpdf = np.cumsum(pdf)
    cumpdf *= 1/cumpdf[-1]

    # input random values
    randv = np.random.uniform(size=nvals)

    # find where random values would go
    idx1 = np.searchsorted(cumpdf, randv)
    # get previous value, avoiding division by zero below
    idx0 = np.where(idx1==0, 0, idx1-1)
    idx1[idx0==0] = 1

    # do linear interpolation in x
    frac1 = (randv - cumpdf[idx0]) / (cumpdf[idx1] - cumpdf[idx0])
    randdist = x[idx0]*(1-frac1) + x[idx1]*frac1

    return randdist