Inverse Transform Sampling

Inverse Transform Sampling is the most basic way to draw a random variable from an arbitrary 1-D distribution $F$: if $U \sim \text{Uniform}(0, 1)$, then $X = F^{-1}(U)$ has distribution $F$. Correctness proof: for any $x$,

the first step holds because $F$ is non-decreasing so $F^{-1}$ preserves order ($F^{-1}(U) \le x \iff U \le F(x)$); the second because $U$ is uniform on $(0,1)$, so the probability equals the interval length $F(x)$; this matches the CDF definition $F_X(x) = P(X \le x)$, hence $X \sim F$.

sampling distributions whose CDF has a closed-form inverse (exponential, Cauchy, Weibull), and as the building block of quantile transforms and copula sampling; interviews often start with "how to draw an exponential sample from a uniform".

one uniform draw plus one function inversion gives exact samples with no rejections and $O(1)$ cost; the hard requirement is an invertible CDF — the normal and gamma distributions have no closed-form inverse (use Box-Muller or numeric inversion such as the inverse erf), and the method is inherently 1-D and does not extend to high dimensions.

Prove inverse transform sampling: derive $P(F^{-1}(U) \le x) = P(U \le F(x)) = F(x)$ and justify each step?

What are the applicability conditions of inverse transform sampling? Why can't it be applied to the normal directly?

Draw $X \sim \text{Exp}(\lambda)$ via inverse transform: derive the explicit $F^{-1}$ and write the algorithm?

Inverse transform sampling vs Box-Muller: differences and when to use each?

How does inverse transform sampling work for discrete distributions (Bernoulli/Poisson)? Complexity?