Let $X_1,...,X_n $ be iid $U(0,\theta)$. Find the UMP to test $H_0: \theta = \theta_0$ versus $H_1: \theta=\theta_1$, for $\theta_1 < \theta_0.$ Obtain the power of the test.

My attempt:

We know that $X_{(n)}$ is sufficient for $\theta$ and its distribution is $$f(x;\theta)=\frac{nx^{n-1}}{\theta^n}I_{(0, \theta)}(x)$$

So, by the Neyman-Person lemma we must have a critical region with the form

$$\{ x; \frac{I_{(0,\theta_0)}(x)}{I_{(0,\theta_1)}(x)}\frac{\theta_1^n}{\theta_0^n} \leq c \}$$

For $0<c<1$

But I can't write it in a better form. What should I do now?

Thanks in advance!


3 Answers 3


UMP test is not unique here.

Since the pdf $f$ has monotone likelihood ratio (MLR) in $X_{(n)}$, by Karlin-Rubin theorem a UMP size $\alpha$ test for testing $H_0:\theta=\theta_0$ against $H_1:\theta<\theta_0$ is

$$\phi_0(x_1,\ldots,x_n)=\begin{cases}1&,\text{ if }x_{(n)}<\theta_0\alpha^{1/n} \\ 0&,\text{ otherwise }\end{cases}$$

Now whenever $\theta_1<\theta_0$, we have $$\frac{f_{\theta_1}(x_1,\ldots,x_n)}{f_{\theta_0}(x_1,\ldots,x_n)}=\begin{cases}\left(\frac{\theta_0}{\theta_1}\right)^n &,\text{ if }0<x_{(n)}\le \theta_1 \\ 0&,\text{ if }\theta_1<x_{(n)}\le \theta_0\end{cases}$$

So by NP lemma, a most powerful level $\alpha$ test $H_0$ against $H_1:\theta=\theta_1(<\theta_0)$ is of the form

$$\phi^*=\begin{cases}0 &,\text{ if }\theta_1<x_{(n)}\le \theta_0 \\\text{any value in }[0,1]&,\text{ otherwise }\end{cases}$$

such that $E_{\theta_0}\phi^*=\alpha$.

This yields a non-randomized MP test for $H_0$ versus $H_1:\theta<\theta_0$, namely

$$ \phi_1(x_1,\ldots,x_n)=\begin{cases}0&,\text{ if }\theta_1<x_{(n)}\le \theta_0 \\ 0 &,\text{ if }\theta_0\alpha^{1/n}<x_{(n)}\le \theta_1 \\ 1 &,\text{ otherwise }\end{cases} $$

The corresponding UMP test is

$$\phi_1(x_1,\ldots,x_n)=\begin{cases}0&,\text{ if }\theta_0\alpha^{1/n}<x_{(n)}\le \theta_0 \\ 1&,\text{ otherwise }\end{cases}$$

Also see Most powerful test of simple vs. simple in $\mathrm{Unif}[0, \theta]$.

  • $\begingroup$ Could you please explain the meaning of the $\phi$ or $\phi^*$ notation? My book doesn't use that notation, and I'm finding it extraordinarily confusing. $\endgroup$ Aug 18, 2021 at 18:59
  • 2
    $\begingroup$ It is a test function (or simply a test); $\phi=1$ indicates rejection of null hypothesis and $\phi=0$ indicates non-rejection. $\endgroup$ Aug 18, 2021 at 19:05
  • $\begingroup$ Ok, thanks! Appreciate the extra clarity. $\endgroup$ Aug 18, 2021 at 19:33
  • $\begingroup$ Sorry, another question: can you please elaborate on how the $\alpha^{1/n}$ arises in the intervals for your $\phi_1$ expressions? Is that simply to enforce $E_{\theta_0}(\phi^*)=\alpha?$ Thanks! $\endgroup$ Aug 19, 2021 at 18:02
  • $\begingroup$ How do you know that $\theta_0\alpha^{1/n}<\theta_1?$ Wouldn't you just want to say that $\phi_a=1$ when $x_{(n)}\le\min(\theta_0\alpha^{1/n},\theta_1)?$ $\endgroup$ Aug 20, 2021 at 14:45

The likelihood-ratio (LR) test is not terribly useful in this situation. Your test can be simplified from your specified critical region by looking at possible regions in which the maximum value can fall. From the ordering in your critical region, it is clear that the p-value function for your test is:

$$p(\boldsymbol{x}) = \begin{cases} \text{undefined} & & & \text{for } \theta_0 < x_{(n)}, \\ 1 & & & \text{for } \theta_1 < x_{(n)} \leqslant \theta_0, \\ (\theta_1 / \theta_0)^n & & & \text{for } 0 \leqslant x_{(n)} \leqslant \theta_1. \\ \end{cases}$$

(In the case where $\theta_0 < x_{(n)}$ both hypotheses are falsified by the data, and your LR statistic is undefined, leading to an undefined p-value.)

We can see that, for any significance level $\alpha < (\theta_1 / \theta_0)^n$ the likelihood-ratio test accepts the null hypothesis under all possible observed outcomes (and is trivially UMP). For any significance level $\alpha > (\theta_1 / \theta_0)^n$, the test rejects the null if and only if $x_{(n)} \leqslant \theta_1$ (and it is again trivially UMP).

The problem with the LR test in this situation is that the LR is either zero or one, and does not have any gradations inside the range $0 \leqslant x_{(n)} \leqslant \theta_1$. This leads to a test with a binary p-value.

A better test to apply here (which does not satisfy the conditions of the Neyman-Pearson lemma, but is also UMP) is to impose an additional evidentiary ordering within the range $0 \leqslant x_{(n)} \leqslant \theta_1$ so that smaller values of $x_{(n)}$ are considered to be greater evidence for the alternative hypothesis. If we add this additional ordering we obtain the smoother p-value function:

$$p(\boldsymbol{x}) = \begin{cases} \text{undefined} & & & \text{for } \theta_0 < x_{(n)}, \\ 1 & & & \text{for } \theta_1 < x_{(n)} \leqslant \theta_0, \\ (x_{(n)} / \theta_0)^n & & & \text{for } 0 \leqslant x_{(n)} \leqslant \theta_1. \\ \end{cases}$$

This latter test has the benefit of avoiding a binary p=value, while maintaining the UMP condition (again trivially). Intuitively, it involves the specification of a lower observed maximum value being more conducive to a lower upper bound in the sampling distribution.

  • $\begingroup$ (+1) The likelihood ratio (when defined) is either 0 or $\left(\frac{\theta_0}{\theta_1}\right)^n$, not that this materially affects the points you make in the 3rd paragraph. A motivation for the additional evidentiary ordering described in the 4th paragraph might be provided by the generalized likelihood-ratio test statistic $\frac{\sup_\theta[f(\vec{x};\theta)]}{f(\vec{x};\theta_0)}=\frac{(1/x_{(n)})^n}{(1/\theta_0)^n}$ $\endgroup$ Sep 5, 2022 at 13:54

I think you might found the answer for this. But this is what i am thinking.

Define $$\frac{\theta_1^n}{\theta_0^n} = T^n$$ then, reject the null hypothesis if $T^n \leq $ c.

by taking the logarithm of both sides , you can simplify this to , reject the null hypothesis if $$log (T) \leq k$$ .

  • 1
    $\begingroup$ The value $T$ is fixed by the hypotheses, and does not depend on the data. So by this reasoning, your test does not depend on the data at all! $\endgroup$
    – Ben
    May 12, 2018 at 1:59
  • $\begingroup$ In Uniform distribution, the likelihood function does not depend on data. Here is a similar kind of that attempted , math.stackexchange.com/questions/1736322/… $\endgroup$ May 14, 2018 at 16:57
  • 1
    $\begingroup$ The uniform in this case depends on the data, through its support. In the link you have provided, the test also depends on the data. $\endgroup$
    – Ben
    May 14, 2018 at 23:48

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