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I'm trying to prove Theorem 10.12 in All of Statistics, which states the following:

Suppose that the size $ \alpha $ test is of the form $$ \text{reject } H_0 \text{ if and only if } T(X^n) \geq c_{\alpha}. $$ Then, $$ \text{p-value} = \sup_{\theta \in \Theta_0} \mathbb{P}_{\theta} (T(X^n) \geq T(x^n)) $$ where $ x^n $ is the observed value of $ X^n $. If $ \Theta_0 = \{\theta_0\} $, then $$ \text{p-value} = \mathbb{P}_{\theta_0} (T(X^n) \geq T(x^n)). $$

Wasserman (the author) defines the p-value as $$ \inf_{\alpha \in (0, 1)} \left\{ \alpha: T(X^n) \in R_{\alpha} \right\} $$ where $ R_\alpha $ is the rejection region for the hypothesis test. (Note that technically I think this should be $ X \in R_{\alpha} $ given that he defines the rejection region as a subset of the domain of the samples not the range of the test statistic.) He defines a size $ \alpha $ test as a test in which $ \sup_{\theta \in \Theta_0} \beta(\theta) = \alpha $ for power function $ \beta(\alpha) = \mathbb{P}_{\alpha}(X \in R) $.

Thus, to try and prove Theorem 10.12, I started by expanding the definition of p-value in terms of the theorem's assumptions, $$ \text{p-value} = \inf_{\alpha \in (0, 1)} \left\{\sup_{\theta \in \Theta_0} \mathbb{P}_{\theta} \left(T(X^n) \geq c_\alpha\right): T(x^n) \geq c_\alpha \right\}. $$

This is where I hit a dead end though. I tried finding a formula for $ c_\alpha $ that doesn't include $ \alpha $ but got stuck having to invert $ \sup $.

It seems like an alternative way to look at this is to view it as saying that the value of $ c_\alpha $ that minimizes $ \sup_{\theta \in \Theta_0} \mathbb{P}_{\theta}(T(X^n) \geq c_\alpha) $ is the observed value of $ T(X^n) $, $ T(x^n) $. But I'm really uncertain how to prove this.

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  • $\begingroup$ Since this is the definition of a p-value, it seems a bit misguided to "prove" it. Instead, just start from where a distribution (like the T-distribution) used to compute p-values comes from. $\endgroup$ Apr 6 '20 at 17:29
  • $\begingroup$ I know this is the definition of a p-value in other texts but here the definition of a p-value is actually the smallest size test that rejects the null and my goal is to prove that that is equivalent to the above definition. $\endgroup$
    – an1lam
    Apr 6 '20 at 17:31
  • $\begingroup$ Rejecting the null is an arbitrary decision. It's made by convention based on a particular p-value. $\endgroup$ Apr 6 '20 at 17:34
  • $\begingroup$ Did you look at what I said above about how Wasserman defines the p-value? He first defines hypothesis tests in terms of hypotheses, power, and size and then defines the p-value as the minimum size test for which we can reject the null hypothesis given some results. $\endgroup$
    – an1lam
    Apr 6 '20 at 17:53
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    $\begingroup$ I think this can be reworded the following way: prove that a $(1-p)\times 100\%$ confidence interval has the hypothesized value as an endpoint. $\endgroup$
    – Dave
    Apr 6 '20 at 18:57
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My solution: we're looking for the smallest $ \alpha $ for which we reject the null hypothesis given our assumptions about the form of the test.

By assumption, for a level $ \alpha $ test, we'll reject the null if and only if $ T(x^n) \geq c_{\alpha} $. Now, note that $ \alpha = \sup_{\theta \in \Theta_0} \mathbb{P}(T(X^n) \geq c_\alpha) $ is a decreasing function of $ c $. Thus, we can recast our goal as maximizing $ c_{\alpha} $ while still rejecting the null.

If we observe $ T(x^n) $, then clearly the maximum value of $ c_{\alpha} $ for which we'll reject the null hypothesis is $ c_{\alpha} = T(x^n) $. A test for which we reject the null if $ T(X^n) \geq T(x^n) $ has size $$ \alpha = \sup_{\theta \in \Theta_0} \mathbb{P}(T(X^n) \geq T(x^n)), $$ so $$ \text{p-value} = \inf_{\alpha \in (0, 1)} \left\{ \alpha: T(X^n) \geq c_\alpha\right\} = \sup_{\theta \in \Theta_0} \mathbb{P}(T(X^n) \geq T(x^n)). $$

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