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📐 AI MathID: inverse-transform

Inverse Transform Sampling

逆变换采样
🎯Core Definition
Inverse Transform Sampling is the most basic way to draw a random variable from an arbitrary 1-D distribution FF: if UUniform(0,1)U \sim \text{Uniform}(0, 1), then X=F1(U)X = F^{-1}(U) has distribution FF. Correctness proof: for any xx,
📌Overview
P(F1(U)x)=P(UF(x))=F(x),P(F^{-1}(U) \le x) = P(U \le F(x)) = F(x),
📌Overview
the first step holds because FF is non-decreasing so F1F^{-1} preserves order (F1(U)x    UF(x)F^{-1}(U) \le x \iff U \le F(x)); the second because UU is uniform on (0,1)(0,1), so the probability equals the interval length F(x)F(x); this matches the CDF definition FX(x)=P(Xx)F_X(x) = P(X \le x), hence XFX \sim F.
💡Use Cases
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".
Key Problems Solved
one uniform draw plus one function inversion gives exact samples with no rejections and O(1)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.
🎯5 High-Frequency Exam Points
1
Prove inverse transform sampling: derive P(F1(U)x)=P(UF(x))=F(x)P(F^{-1}(U) \le x) = P(U \le F(x)) = F(x) and justify each step?
2
What are the applicability conditions of inverse transform sampling? Why can't it be applied to the normal directly?
3
Draw XExp(λ)X \sim \text{Exp}(\lambda) via inverse transform: derive the explicit F1F^{-1} and write the algorithm?
4
Inverse transform sampling vs Box-Muller: differences and when to use each?
5
How does inverse transform sampling work for discrete distributions (Bernoulli/Poisson)? Complexity?
Updated 2026-08-12
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