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Say $X \sim \mathrm{Unif}[0, \theta]$. Denote the observations as $x_i$ $(i=1, \cdots, n)$.

Show that any test $\phi$ that satisfies the following two conditions is most powerful test of level $\alpha$ for $H_0 : \theta = \theta_0$ vs. $H_1 : \theta = \theta_1$ $(\theta_1 > \theta_0)$.

  • $\phi^{*}(x_1, \cdots, x_n) = 1 \ \ (\mathrm{if} \max\{x_1, \cdots, x_n\}) > \theta_0$

  • $\mathbb{E}_{\theta_0} \phi^{*}(X) = \alpha$


Try

By Neyman-Pearson lemma, any test that satisfies

  • $\phi^{*}(x_1, \cdots, x_n) = \begin{cases} 1 & (\mathcal{L}(\theta_1 | x_1, \cdots, x_n)/\mathcal{L}(\theta_0 | x_1, \cdots, x_n) > k) \\ \gamma & (\mathcal{L}(\theta_1 | x_1, \cdots, x_n)/\mathcal{L}(\theta_0 | x_1, \cdots, x_n) = k) \\ 0 & (\mathcal{L}(\theta_1 | x_1, \cdots, x_n)/\mathcal{L}(\theta_0 | x_1, \cdots, x_n) < k) \end{cases} $

(for some $\gamma \in [0,1]$)

  • $\mathbb{E}_{\theta_0} \phi^{*}(X) = \alpha$

is an MP test for $H_0$ vs. $H_1$.

Since

$$ \begin{aligned} \mathcal{L}(\theta_1 | x_1, \cdots, x_n)/\mathcal{L}(\theta_0 | x_1, \cdots, x_n) &= (\theta_0 / \theta_1)^n I(\max x_i \le \theta_1)/I(\max x_i \le \theta_0) \\ &= \begin{cases} (\theta_0/\theta_1)^n & (\max x_i > \theta_0) \\ \infty & (\max x_i \le \theta_0) \end{cases} \end{aligned} $$

we have

$$ \phi^{*}(x_1, \cdots, x_n) = \begin{cases}1 & (\max x_i > \theta_0) \\ \alpha & (\max x_i \le \theta_0) \end{cases} $$

is an MP test for $H_0$ vs. $H_1$.

But this does not mean ANY test that satisfies

  • $\phi^{*}(x_1, \cdots, x_n) = 1 \ \ (\mathrm{if} \max\{x_1, \cdots, x_n\}) > \theta_0$

  • $\mathbb{E}_{\theta_0} \phi^{*}(X) = \alpha$

is an MP test, but rather just provides an example of MP test.

Is there anyone to help me out?

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  • $\begingroup$ You might want to check the inequality signs when you say $\begin{cases} (\theta_0/\theta_1)^n & (\max x_i > \theta_0) \\ \infty & (\max x_i \le \theta_0) \end{cases}$ as I suspect you intended these the other way round. It does not affect the rest of your argument $\endgroup$ – Henry Nov 30 '18 at 1:44
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The logic is:

  • Any test which satisfies the two conditions has the same significance level $\alpha$ and power $1-\left(\frac{\theta_0}{\theta_1}\right)^n +\alpha\left(\frac{\theta_0}{\theta_1}\right)^n$ as the particular test you found; the

  • Given that your test is most powerful of all tests with significance level $\alpha$, all the other tests satisfying the two conditions are also most powerful with significance level $\alpha$

If you want to check, consider the significance level and power of these two deterministic tests

  • $\phi^{*}(x_1, \cdots, x_n) = \begin{cases} 1 & \text{ when }\max x_i > \theta_0 \sqrt[n]{1-\alpha} \\ 0 & \text{ when }\max x_i \le \theta_0 \sqrt[n]{1-\alpha} \end{cases}$
  • $\phi^{*}(x_1, \cdots, x_n) = \begin{cases} 1 & \text{ when }\max x_i > \theta_0 \\ 0 & \text{ when }\theta_0 \sqrt[n]{\alpha} \lt \max x_i \le \theta_0 \\ 1 & \text{ when }\max x_i \le \theta_0 \sqrt[n]{\alpha} \end{cases}$
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