Prove the p-value is uniform under the null

Question

Consider a test with continuous test statistic $T$ , rejecting for large values, with p-value $P = 1 - F_0(T)$ where $F_0$ is the null CDF of $T$ .

Prove that $P \sim \text{Uniform}(0,1)$ when the null is true, explain what happens under the alternative, and describe how this fact is used in practice.

Accepted Answer

The proof. Assume F0 is continuous and strictly increasing on the support of T (so F0^{-1} exists). Under H0, for any u \in (0,1): \mathbb{P}(P \leq u) = \mathbb{P}\big(1 - F0(T) \leq u\big) = \mathbb{P}\big(F0(T) \geq 1 - u\big) = \mathbb{P}\big(T \geq F0^{-1}(1-u)\big) = 1 - F0\big(F0^{-1}(1-u)\big) = u. That is the uniform CDF, so P \sim 	ext{U}(0,1). This is the probability integral transform in action: plugging any continuous random variable into its own CDF yields a uniform. (For discrete statistics, P is stochastically larger than uniform, such tests are conservative.) An immediate co…

Prove the p-value is uniform under the null

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