如何从R中的双变量已知PDF生成随机变量? [英] How to generate random variables from a bivariate known PDF in R?

问题描述

我在DX x DY矩形区域中具有此二元概率密度函数:

I have this bivariate probability density function in a DX x DY rectangular region:

链接到我的pdf

我正在使用R.如何按照该pdf分布在矩形内生成随机(x，y)点?

I am using R. How can I generate random (x,y) points within the rectangle following this pdf distribution?

我已经阅读了许多有关逆变换采样"的答案，但是我没有单变量pdf.我已经见过此方法，但是看起来很乏味并且很难实现.

I have read many answers regarding "inverse transform sampling", but I don't have a univariate pdf. I have seen this method, but looks tedious and extremely difficult to implement.

提前谢谢！

推荐答案

这似乎更多是统计信息，而不是编程问题.无论如何，拒收抽样应适合您的情况.您可以在维基百科中阅读详细信息.以下函数执行拒绝采样(n ...个样本数； pdf ...概率密度函数； maxval ... pdf可产生的最大值(或更大)； xlim，ylim ...边界框，其中密度函数产生值大于零):

This seems to be more of a statistical than a programming question. Anyway, rejection sampling should work in your case. You can read up the details at wikipedia. The following function performs rejection sampling (n...number of samples; pdf...probability density function; maxval...maximum (or bigger) value your pdf can yield; xlim, ylim... boundary box wherein density function yields values greater zero):

reject.sample.2d <- function(n,pdf,maxval,xlim,ylim)
{
  smpl <- data.frame(x=numeric(n),y=numeric(n))
  i <- 0
  while (i<n)
  {
    xval <- runif(1,xlim[1],xlim[2])
    yval <- runif(1,ylim[1],ylim[2])
    if (runif(1)<pdf(xval,yval)/maxval)
    {
      i <- i+1
      smpl[i,] <- c(xval,yval)
    }
  }
  return(smpl)
}

例如，它可以用于二维正态分布

For instance, it can be used for the 2d normal distribution

mydens <- function(x,y)
{
  dnorm(x)*dnorm(y)
}

以这种方式(此处定义的2d正态分布的最大值为x = 0，y = 0且小于0.16):

in that way (the maximum value of the here defined 2d normal distribution is at x=0,y=0 and a bit smaller than 0.16):

res <- reject.sample.2d(5000,mydens,0.16,c(-5,5),c(-5,5))

一些检查:

> sd(res[["x"]])
[1] 1.015413
> sd(res[["y"]])
[1] 0.9981738

> shapiro.test(res[["x"]])

    Shapiro-Wilk normality test

data:  res[["x"]]
W = 0.9995, p-value = 0.1603

> shapiro.test(res[["y"]])

    Shapiro-Wilk normality test

data:  res[["y"]]
W = 0.9997, p-value = 0.8304

p.s. 您也可以使用逆变换采样.计算沿x轴的边际分布.使用此分布进行逆变换采样以获得x值.对于此x值，请使用条件概率分布(给定获得的x，y的概率)，然后再次对i值使用逆变换采样.

p.s. You can also use inverse transform sampling. Calculate the marginal distribution along i.e. the x-axis. Use this distribution for inverse transform sampling to get an x-value. For this x-value use the conditional probability distribution (probability for y given the obtained x) and use inverse transform sampling again, now for the y-value.

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