如何从R中的pareto密度函数生成随机数据？

Question

I am trying to generate data from given pareto density in R.

Pareto density: F(x) = |X|^(-3) * 1 |x|>1

I know that I need to use rpareto function from actuar library, but I am not sure how should I transform given pareto density into parameters.

Answer 1

The Pareto distribution is simple enough to allow for the transformation method (aka Inverse Transform Sampling ). The CDF of the Pareto distribution is

$$F(x) = \\int_{x_{min}}^xf(x')\\,dx' = 1- \\left(\\frac{x_{min}}{x}\\right^k$$

which can be analytically inverted to

$$F^{-1}(y) = x_{min} \\cdot (1-y)^{-1/k}$$

Thus, if you transform random numbers uniformly distributed on $(0,1)$ with $F^{-1}$, you obtain Pareto distributed random numbers.

Edit: Oh sorry, apparently Latex-Code is not supported here. Here is, for your convenince, the R code:

k <- 5      # parameter k of the Pareto distribution
x.min <- 2  # cutoff point of Pareto distribution
N <- 500    # number of random points
x.random <- x.min*(1-runif(N))^(-1/k)

And here is the practical demonstration that it works:

h <- hist(x.random, freq=FALSE, plot=TRUE)
x <- seq(x.min, h$breaks[length(h$breaks)], by=0.01)
lines(x, k*x.min^k/x^(k+1), col="red")

plot for k=5 and x.min=2

Answer 2

Using inverse sampling method, formulas from Pareto Distribution

Here is updated version with mean and sd computed

rpar <- function(n, xm, a) {
    v <- runif(n)
    xm / v^(1.0/a)
}

rpar_mean <- function(xm, a) {
    result <- 1/0 # Inf
    if (a > 1.0)
        result <- a*xm/(a - 1.0)
    result
}

rpar_var <- function(xm, a) {
    result <- 1/0 # Inf
    if (a > 2.0)
        result <- xm*xm*a/((a - 1.0)^2*(a - 2.0))
    result
}

set.seed(54122345)

xm = 1.0
a  = 3.0

q <- rpar(10000, xm, a)
print(mean(q))
print(rpar_mean(xm, a))

print(sd(q))
print(sqrt(rpar_var(xm, a)))