Sampling from the uniform and the inverse transform

Following Exercise 2.2. from Introducing Monte Carlo Methods with R by Robert & Casella.

Given

$$u = F(x) = \frac{1}{1 + e^{\frac{-(x-\mu)}{\beta}}},$$

we rearrange for $x$,

$$x = F^{-1}(u) = \mu - \beta \ln(1/u - 1),$$

with $u \sim U(0, 1)$.

Both sampling from $u$ and applying the inverse function and sampling from the logistic distribution yield similar results.

	set.seed(42)
	antiF <- function(u, mu, beta) {mu - beta * log(1/u - 1)}
	n = 10000
	mu = 5
	beta = 2
	s1 = antiF(runif(n), mu, beta)
	s2 = rlogis(n, mu, beta)

view raw robert-casella-ex-2.2.R hosted with ❤ by GitHub

We visualise the similarity,

Neato!