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See here for a PDF version of this vignette.

Prerequisites

This document assumes basic familiarity with probability theory.

Overview

Inverse transform sampling is a method for generating random numbers from any probability distribution by using its inverse cumulative distribution, \(F^{-1}(x)\). Recall that the cumulative distribution for a random variable \(X\) is \(F_X(x) = P(X \leq x)\). In what follows, we assume that our computer can, on demand, generate independent realizations of a random variable \(U\) uniformly distributed on \([0,1]\).

Algorithm

Continuous distributions

Assume we want to generate a random variable \(X\) with cumulative distribution function (CDF) \(F_X\). The inverse transform sampling algorithm is simple:

Generate \(U \sim \mathrm{Unif}(0,1)\).
Let \(X = F_X^{-1}(U)\).

Then \(X\) will follow the distribution governed by the CDF \(F_X\), which was our desired result.

Note that this algorithm works in general but is not always practical. For example, inverting \(F_X\) is easy if \(X\) is an exponential random variable, but it is harder if \(X\) is normal random variable.

Discrete distributions

Now we will consider the discrete version of the inverse transform method. Assume that \(X\) is a discrete random variable such that \(P(X = x_i) = p_i\). The algorithm proceeds as follows:

Generate \(U \sim \text{Unif}(0,1)\).
Determine the index \(k\) such that \(\sum_{j=1}^{k-1} p_j \leq U < \sum_{j=1}^k p_j\), and return \(X = x_k\).

(Notice that the second step requires a search. )

Assume our random variable \(X\) takes on one of \(K\) values with probabilities \(\{p_1, \ldots, p_K\}\). We implement the algorithm below, assuming these probabilities are stored in a vector called “pvec”.

discrete_inv_transform_sample <- function (pvec) {
  K <- length(pvec)
  u <- runif(1)
  if (u <= pvec[1])
    return(1)
  for (k in seq(2,K)) {
    if(sum(pvec[seq(1,k-1)]) < u && u <= sum(pvec[seq(1,k)]))
      return(k)
  }
}

Note that this is this an inefficient implementation given here for pedagogical purposes.

Continuous example: exponential distribution

Assume \(Y\) is an exponential random variable with rate parameter \(\lambda = 2\). Recall that the probability density function is \(p(y) = 2e^{-2y}\), for \(y > 0\). First, we derive the CDF: \[ F_Y(x) = P(Y\leq x) = \int_0^x 2e^{-2y} dy = 1 - e^{-2x}. \] Solving for the inverse CDF, we get that \[ F_Y^{-1}(y) = \textstyle -\frac{1}{2}\log(1-y). \]

Using our algorithm above, we first generate \(U \sim \text{Unif}(0,1)\), then set \(X = -\log(1-U)/2\). We do this in the R code below and compare the histogram of our samples with the true density of \(Y\).

set.seed(1)
num_samples <- 10000
u <-  runif(num_samples)
x <- -log(1-u)/2
hist(x,n = 32,freq = FALSE,xlab = 'X',main = "")
curve(dexp(x,rate = 2),0,3,lwd = 2,add = TRUE)

Version	Author	Date
3b61f33	Peter Carbonetto	2026-01-05

Indeed, the random draws appear are following the intended distribution.

Discrete example

Let’s assume we want to simulate a discrete random variable \(X\) that follows the following distribution:

\(k\)	\(\Pr(X = k)\)
1	0.1
2	0.4
3	0.2
4	0.3

Below we simulate from this distribution using the “discrete_inv_transform_sample” function above, and plot both the true probability vector, and the empirical proportions from our simulation.

par(mfcol = c(1,2))
num_samples <- 1000
pvec        <- c(0.1,0.4,0.2,0.3)
names(pvec) <- 1:4
samples     <- rep(0,num_samples)
for(i in seq_len(num_samples)) {
  samples[i] <- discrete_inv_transform_sample(pvec)
}
barplot(pvec,main = "true pdf")
barplot(table(samples),main = "empirical pdf")

Version	Author	Date
3b61f33	Peter Carbonetto	2026-01-05

Again, the plot supports our claim that we are drawing from the correct probability distribution.

sessionInfo()
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Inverse transform sampling

Matt Bonakdarpour

January 5, 2025