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Hypothesis Testing and the One-to-One Correspondence of Confidence Sets 📂Mathematical Statistics

Hypothesis Testing and the One-to-One Correspondence of Confidence Sets

Theorem

Let’s assume we have parameter space Θ\Theta and space X\mathcal{X} given.

Explanation

To briefly summarize the motivation behind this theorem, it is as follows: θ0C(x)    xA(θ0) \theta_{0} \in C \left( \mathbf{x} \right) \iff \mathbf{x} \in A \left( \theta_{0} \right)

Proof 1

(    )\left( \implies \right)

Since A(θ0)A \left( \theta_{0} \right) is the rejection region of level α\alpha, Pθ0(XA(θ0))αPθ0(XA(θ0))1α \begin{align*} P_{\theta_{0}} \left( \mathbf{X} \notin A \left( \theta_{0} \right) \right) \le & \alpha \\ P_{\theta_{0}} \left( \mathbf{X} \in A \left( \theta_{0} \right) \right) \ge & 1 - \alpha \end{align*} As it holds for all θ0\theta_{0} given the assumption, we can write it as θ\theta, and since we defined C(x)={θ0:xA(θ0)}C \left( \mathbf{x} \right) = \left\{ \theta_{0} : \mathbf{x} \in A \left( \theta_{0} \right) \right\}, the coverage probability of C(X)C \left( \mathbf{X} \right) is Pθ(XC(X))=Pθ(XA(θ))1α P_{\theta} \left( \mathbf{X} \in C \left( \mathbf{X} \right) \right) = P_{\theta} \left( \mathbf{X} \in A \left( \theta \right) \right) \ge 1 - \alpha In other words, C(X)C \left( \mathbf{X} \right) is a 1α1-\alpha confidence set.


(    )\left( \impliedby \right)

The probability of a type I error for A(θ0)A \left( \theta_{0} \right) in H0:θ=θ0H_{0} : \theta = \theta_{0} is Pθ0(XA(θ0))=Pθ0(θ0C(X))α P_{\theta_{0}} \left( \mathbf{X} \notin A \left( \theta_{0} \right) \right) = P_{\theta_{0}} \left( \theta_{0} \notin C \left( \mathbf{X} \right) \right) \le \alpha Therefore, it is a level α\alpha hypothesis test.


  1. Casella. (2001). Statistical Inference(2nd Edition): p422. ↩︎