Inverse transform sampling

Remarks

The key fact is that for $u \in (0, 1)$ and $x \in \R$, $F^{-1}(u) \le x$ if and only if $u \le F(x)$; this uses right-continuity of $F$, which makes the infimum attained. Hence $P(X \le x) = P(U \le F(x)) = F(x)$. Example: the exponential distribution with rate $\lambda > 0$ has $F(x) = 1 - e^{-\lambda x}$ for $x \ge 0$, so $F^{-1}(u) = -\ln(1 - u)/\lambda$; since $1 - U$ is also uniform, $-\ln(U)/\lambda$ is exponential too. Example: for a discrete distribution with values $x_1 < x_2 < \cdots$ and probabilities $p_1, p_2, \ldots$, the method returns $x_k$ when $p_1 + \cdots + p_{k-1} < U \le p_1 + \cdots + p_k$. The generalized inverse is needed: when $F$ has jumps or flat parts, the ordinary inverse does not exist. A converse holds when $F$ is continuous: then $F(X)$ is uniform on $(0, 1)$, the probability integral transform.

Going deeper

For a student

The method is exact and uses one uniform number per sample, so it is the first choice whenever $F^{-1}$ has a closed form or a fast accurate approximation. For the normal distribution there is no closed form, and one uses numerical approximations of $F^{-1}$ or other methods such as the Box–Muller transform or the ziggurat algorithm. Being monotone in $U$, it preserves the order of the uniform inputs, which is useful for antithetic variates, for quasi-Monte Carlo and for coupling arguments. It also explains copulas: any multivariate distribution is obtained from its copula by applying the marginal quantile functions coordinatewise.