Question

Consider the situation described immediately before Eq. (9.1.12). Prove that the expression (9.1.12) equals the smallestα₀such that we would reject H₀ at level of significanceα₀.

Short answer

\({P_\theta }(T \ge t)\)is non-increasing in \(t\), thus, the smallest \({\alpha _0}\)that rejects the null hypothesis is \(\mathop {\sup }\limits_{\theta \in {\Omega _0}} {P_\theta }(T \ge t)\).

Step-by-step solution

Definition of the p-value

A p-value is defined as the smallest α₀ such that the null hypothesis is rejected at level α₀ with the observed data.

Proving the function

The test δ_t rejects the null for large values of a statistics T. Now, P_θ( T≥t) is non-increasing in t for any statistic T. Thus, the smallest probability under the null hypothesis thatδ_t rejects the null hypothesis occurs at

\begin{aligned}Sup\underset{_{\theta e}\Omega _{0}}{P_{\theta}}(T\geq t)\end{aligned}

This quantity equals its p-value by its very definition, so our argument is complete.