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when I read Casella's Statistical Inference, the exercise 8.51 is: enter image description here

where (8.3.9) is $p(x)=\sup_{\theta\in\Theta_0}P_\theta(W(X)\geq W(x))$, and $W(X)$ is a statistic s.t. the large values give evidence of the alternative hypothesis.

and the solution is enter image description here

My questions are:

Why $W(x)<c_\alpha$? Does the author implicitly used the assumption that $\arg_\theta\sup_{\theta\in\Theta_0}P_\theta(W(X)\geq c_\alpha)=\arg_\theta\sup_{\theta\in\Theta_0}P_\theta(W(X)\geq W(x))$. Why is it true? And do they assume that $W$ is a statistic s.t. the large values give evidence of the alternative hypothesis, which is not mentioned in the problem.

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  • $\begingroup$ You seem to have skipped over parts of the question in your reading of it. In particular, the rejection region "$\{\mathrm{x}:W(\mathrm{x})\ge c_\alpha\}$" explicitly says that large values of $W$ are evidence against the null. That leaves you only with the first question, the reason why $W(\mathrm{x})\lt c_\alpha$. That follows directly from the definition of $\sup$ and axioms of probability; no additional hidden assumptions were used. Possibly, then, a more careful re-reading of the question and its answer will address all the issues you bring up. $\endgroup$
    – whuber
    Mar 22, 2015 at 18:11

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Thanks for whuber's explanation on $\{x:W(x)\geq c_\alpha\}$. To show $W(x)<c_\alpha$, we can use contradiction. Given $\sup_\theta P(W(X)\geq c_\alpha)<\sup_\theta P(W(X)\geq W(x))$, assume that $W(x)\geq c_\alpha$, then $\text{LHS}=\sup_\theta[P(W(X)\geq W(x))+P(c_\alpha\leq W(X)<W(x))]\geq \sup_\theta[P(W(X)\geq W(x))]=\text{RHS}$. Contradict! Hence, $W(x)<c_\alpha$.

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