问从R中未知密度的分位数生成随机样本

EN

f(x)     = your density function, 
F(x)     = cdf of f(x), and 
F.inv(y) = inverse cdf of f(x).

f <- function(x) {((x-1)^2) * exp(-(x^3/3-2*x^2/2+x))}
F <- function(x) {integrate(f,0,x)$value}
F <- Vectorize(F)

F.inv <- function(y){uniroot(function(x){F(x)-y},interval=c(0,10))$root}
F.inv <- Vectorize(F.inv)

x <- seq(0,5,length.out=1000)
y <- seq(0,1,length.out=1000)

par(mfrow=c(1,3))
plot(x,f(x),type="l",main="f(x)")
plot(x,F(x),type="l",main="CDF of f(x)")
plot(y,F.inv(y),type="l",main="Inverse CDF of f(x)")

X <- runif(1000,0,1)   # random sample from U[0,1]
Z <- F.inv(X)

par(mfrow=c(1,2))
plot(x,f(x),type="l",main="Density function")
hist(Z, breaks=20, xlim=c(0,5))

   drawF <- function(n) {  
     f <- function(x) ((x-1)^2) * exp(-(x^3/3-2*x^2/2+x))
     x <- runif(n, 0 ,4)
     z <- runif(n) 
     subset(x, z < f(x)) # Rejection
   }

sample(x, size, replace = FALSE, prob = NULL)


Arguments

x       Either a vector of one or more elements from which to choose, or a positive integer. See ‘Details.’

n    a positive number, the number of items to choose from. See ‘Details.’

size    a non-negative integer giving the number of items to choose.

replace    Should sampling be with replacement?

prob    A vector of probability weights for obtaining the elements of the vector being sampled.