The probability integral transform and its inverse

Question

Let X be a continuous random variable with strictly increasing CDF F. Prove that U = F(X) \sim 	ext{Uniform}(0,1), then show how to reverse it to generate samples of any target distribution from uniform random numbers.

Accepted Answer

Forward direction (PIT). For u \in (0,1), since F is strictly increasing and invertible, \mathbb{P}(U \le u) = \mathbb{P}\big(F(X) \le u\big) = \mathbb{P}\big(X \le F^{-1}(u)\big) = F\big(F^{-1}(u)\big) = u. That is exactly the CDF of a Uniform(0,1), so U = F(X) \sim 	ext{Uniform}(0,1). Feeding any continuous variable into its own CDF flattens it to a uniform. Reverse direction (inverse-transform sampling). Run the same map backward. Let U \sim 	ext{Uniform}(0,1) and define Y = F^{-1}(U) for a target CDF F. Then \mathbb{P}(Y \le y) = \mathbb{P}\big(F^{-1}(U) \le y\big) = \mathbb{P}\big(U \l…

The probability integral transform and its inverse

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