# Inverting a hypothesis test: nitpicky detail

First timer here. I'm reviewing a section of Casella and Berger on inverting a hypothesis test to get a Confidence Interval (CI), and came across a proof which I believe is missing a tiny piece. Here's the statement of the theorem verbatim from CB:

Theorem: For each $\theta_0 \in \Theta$, let $A(\theta_0)$ be the acceptance region of a level $\alpha$ test of $H_0: \theta = \theta_0$. For each $\boldsymbol{x} \in \mathcal{X}$, define a set $C(\boldsymbol{x})$ in the parameter space by $$C(\boldsymbol{x}) = \{\theta_0: \boldsymbol{x} \in A(\theta_0)\}$$ then the random set $C(\boldsymbol{X})$ is a $1-\alpha$ confidence set.

Proof: We want to show that the confidence coefficient is equal to $1-\alpha$. Fix some $\theta_0 \in \Theta$ and define $R(\theta_0)$ as the rejection region of the aforementioned level $\alpha$ test of $H_0: \theta = \theta_0$. Then:

\begin{align*} P_{\theta_0}(\theta_0 \in C(\boldsymbol{X})) &= P_{\theta_0}(\boldsymbol{X} \in A(\theta_0))\\ &= 1 - P_{\theta_0}(\boldsymbol{X} \in R(\theta_0))\\ &\geq 1-\alpha \end{align*} Where the last step follows from the fact that the hypothesis test is level $\alpha$. This is true for any $\theta_0 \in \Theta$, so thus $C(\boldsymbol{X})$ is a $1-\alpha$ confidence set.

Qualm: My problem with this proof is that we have shown that all the coverage probabilities are $\geq 1-\alpha$, but we have not shown that the confidence coefficient is equal to $1-\alpha$ (confidence coefficient = infimum of the coverage probabilities). More precisely, the the definition of a $1-\alpha$ confidence set $C(\boldsymbol{X})$ is: $$\inf_{\theta} P_{\theta}(\theta \in C(\boldsymbol{X})) = 1-\alpha$$ It appears to me that the conclusion of the proof only shows that $$\inf_{\theta} P_{\theta}(\theta \in C(\boldsymbol{X})) \geq 1-\alpha$$ I suppose this is a "better" conclusion, but still.

This is a very nitpicky detail. Nevertheless I'm curious what I'm missing. CB don't mention anything else in their presentation, so I figure I'm just having a brain freeze. Any pointers?

ANSWERED: After some deliberation with MansT below, we found it to be a typo in Casella and Berger. The correct theorem statement should read: $$\textrm{...then the random set } C(\boldsymbol{X}) \textrm{ has confidence coefficient } \geq 1-\alpha$$

• If you define your CI's as having at least coverage $1-\alpha$ everything works. It's often impossible to get exactly $1-\alpha$ (consider a binomial, for example - the sample is discrete so the distribution of possible values that the MLE of the continuous parameter takes is discrete). Commented Jun 5, 2013 at 22:38
• Yeah, you're absolutely right. But it's not defined that way in Casella and Berger. Given how classic of a text it is, I'm convinced that I'm just missing something... Commented Jun 5, 2013 at 22:43

As Glen_b pointed out in a comment, for discrete distributions $P_\theta(\theta\in C(\mathbf{X}))$ won't be exactly equal to $1-\alpha$ for all $\theta$. You are however correct in that the classic defintion of a confidence interval is that $\inf_{\theta\in\Theta}P_\theta(\theta\in C(\mathbf{X}))=1-\alpha$ and not just that $\inf_{\theta\in\Theta}P_\theta(\theta\in C(\mathbf{X}))\geq 1-\alpha$.
The tiny piece that is missing is the following. $R(\theta)$ is the rejection region of a level $\alpha$ test. This means that $$P_{\theta}(\mathbf{X}\in R(\theta))\leq \alpha\qquad\forall \theta\in\Theta,$$ a fact that is used in the proof. But, by definition, it also means that $$\sup_{\theta\in\Theta} P_{\theta}(\mathbf{X}\in R(\theta))=\alpha.$$ Consequently, it follows that $$\inf_{\theta\in\Theta}P_\theta(\theta\in C(\mathbf{X}))=\inf_{\theta\in\Theta}\Big(1- P_{\theta}(\mathbf{X}\in R(\theta))\Big)=1-\sup_{\theta\in\Theta} P_{\theta}(\mathbf{X}\in R(\theta))=1-\alpha,$$ just as we were hoping.
• @gogurt: Right! You need a size alpha test in order to have infimum coverage probability $1-\alpha$ (as can be seen from the last line in my answer)... so it's a typo of some sorts - probably the authors getting their own terminology confused :) Commented Jun 6, 2013 at 17:37