使用逆采样从分布函数生成随机变量 [英] Generate random variables from a distribution function using inverse sampling

问题描述

我有一个特定的密度函数，我想知道密度函数的表达式以生成随机变量.

I have a specific density function and I want to generate random variables knowing the expression of the density function.

例如，密度函数为:

df=function(x) { - ((-a1/a2)*exp((x-a3)/a2))/(1+exp((x-a3)/a2))^2 }

我想从这个表达式中生成1000个具有相同分布的随机元素.

From this expression I want to generate 1000 random elements with the same distribution.

我知道我应该使用逆采样方法.为此，我使用PDF的CDF函数，其计算如下:

I know I should use the inverse sampling method. For this, I use the CDF function of my PDF which is calculated as follows:

cdf=function(x) { 1 - a1/(1+exp((x-a3)/a2))

这个想法是生成均匀分布的样本，然后使用我的CDF函数将其映射以获得逆映射.像这样:

The idea is to generate uniformly distributed samples and then map them with my CDF functions to get an inverse mapping. Something like this:

random.generator<-function(n) sapply(runif(n),cdf) 

，然后使用所需数量的随机变量来调用它来生成.

and then call it with the desired number of random variables to generate.

random.generator(1000) 

这种方法正确吗?

推荐答案

第一步是获取cdf函数的逆函数，在这种情况下，可以通过简单的算术来完成:

The first step is to take the inverse of your cdf function, which in this case can be done with simple arithmetic:

invcdf <- function(y) a2 * log(a1/(1-y) - 1) + a3

现在，您要使用标准均匀分布的随机变量调用逆CDF进行采样:

Now you want to call the inverse cdf with standard uniformly distributed random variables to sample:

set.seed(144)
a1 <- 1 ; a2 <- 2 ; a3 <- 3
invcdf(runif(10))
#  [1] -2.913663  4.761196  4.955712  3.007925  1.472119  4.138772 -3.568288
#  [8]  4.973643 -1.949684  6.061130

这是10000个模拟值的直方图:

This is a histogram of 10000 simulated values:

hist(invcdf(runif(10000)))

这是pdf的图:

x <- seq(-20, 20, by=.01)
plot(x, df(x))

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