10.1 The Probability Integral Transform
Theorem 10.1.1. Let \(X\) have a continuous, strictly increasing distribution function \(F\). Then \[U = F(X) \ \sim \ \text {UNIF}(0,1).\] Conversely, if \(U\sim \text {UNIF}(0,1)\) then \(X = F^{-1}(U)\) has distribution function \(F\).
Proof. For \(0<u<1\), using that \(F\) is increasing and therefore invertible, \[P(U\leq u) = P\left (F(X)\leq u\right ) = P\left (X \leq F^{-1}(u)\right ) = F\left (F^{-1}(u)\right ) = u ,\] which is the distribution function of UNIF\((0,1)\). For the converse, set \(X=F^{-1}(U)\); then \[P(X\leq x) = P\left (F^{-1}(U)\leq x\right ) = P\left (U\leq F(x)\right ) = F(x).\] □
Note. The result is the reason a single uniform generator suffices for everything. It also explains why \(p\)-values are uniform under the null hypothesis — a \(p\)-value is the distribution function of the test statistic evaluated at the statistic itself, which is exactly \(F(X)\).
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