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This question is quite simple. We have a random sample $X_1, X_2, ..., X_n$ from $U(\theta, \theta + 1)$ and we want to test $H_0: \theta=0$ vs $H_0: \theta=\theta_1$ for some $0 < \theta_1 < 1$. We want to use Neyman-Pearson's lemma to find the most powerful test. The likelihood-ratio is

$\lambda = \frac{L_0}{L_1} = \frac{L(\theta_0 , x_1,x_2,...,x_n)}{L(\theta_1 , x_1,x_2,...,x_n)} = \frac{I_{(0, \infty)}(y_1)I_{(-\infty,1)}(y_n)}{I_{(\theta_1, \infty)}(y_1)I_{(-\infty,\theta_1 + 1)}(y_n)}$, where $y_1 = min(x_1,...,x_n)$ and $y_n = max(x_1,...,x_n)$.

According to Neyman-Pearson, the most powerful test is $\Lambda \leqslant k$ for some $k$. If we want the test to have size $\alpha$ then we have to find $k$ such that $P_{\theta = 0}(\Lambda \leqslant k) = \alpha$.

If $\theta = 0$, $\Lambda$ equals just $\frac{1}{I_{(\theta_1, \infty)}(y_1)}$ so now we have to find $k$ such that $P_{\theta = 0}(\frac{1}{I_{(\theta_1, \infty)}(y_1)} \leqslant k) = \alpha$.

I don't know how to find such a $k$ and feel like I may be misunderstanding something.

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  • $\begingroup$ You will need to explore the possibility of data for which the likelihood under one hypothesis is $1$ and under the other hypothesis is $0.$ In these cases your expression for $\Lambda$ is undefined, so you had better not rely on it. $\endgroup$
    – whuber
    May 6, 2019 at 15:14
  • $\begingroup$ So Neyman-Pearson's lemma is inapplicable in this situation? $\endgroup$ May 6, 2019 at 16:40
  • $\begingroup$ It has to be carefully interpreted in this situation. Consider rewriting the event $\Lambda \le k$ as $L_0 \le kL_1.$ $\endgroup$
    – whuber
    May 6, 2019 at 16:45
  • $\begingroup$ Hi: This is not my field but ( my memory tells me ) that the uniform doesn't satisfy the regularity conditions for a standard Neyman Pearson approach. You have to use more intuition and I'm not even sure if the "heuristic" test is UMP. Hint: Think about what happens if you took a sample and the maximum value of the observations in such a sample was 1.9999. what would that tell you about $\theta$. $\endgroup$
    – mlofton
    May 6, 2019 at 17:01
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    $\begingroup$ I don't think the theorem asserts you can create a test of exactly the intended size--but you definitely can control it. $\endgroup$
    – whuber
    May 7, 2019 at 13:27

1 Answer 1

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The likelihood ratio requires a bit more careful study here.

Pdf of the sample $X_1,\ldots,X_n$ is $$f_{\theta}(\mathbf x)=\prod_{i=1}^n \mathbf1_{\theta<x_i<\theta+1}=\mathbf1_{\theta<x_{(1)},x_{(n)}<\theta+1}\quad,\,\theta\in\mathbb R$$

I would define the ratio as $\Lambda=f_{H_1}/f_{H_0}$, for which I get the following cases:

\begin{align} \Lambda(\mathbf x)=\frac{f_{H_1}(\mathbf x)}{f_{H_0}(\mathbf x)}&=\frac{\mathbf1_{\theta_1<x_{(1)},x_{(n)}<\theta_1+1}}{\mathbf1_{0<x_{(1)},x_{(n)}<1}} \\\\&=\begin{cases}1&,\text{ if }\theta_1<x_{(1)},x_{(n)}<1 \\ \infty &,\text{ if }x_{(1)}>\theta_1,1<x_{(n)}<\theta_1+1 \\ 0&,\text{ if }0<x_{(1)}<\theta_1,x_{(n)}<1\end{cases} \end{align}

By N-P lemma, you are to reject $H_0$ for large values of $\Lambda$. Can you give it a try now?


I found a similar exercise here that asks to prove that a UMP test for testing $H_0:\theta\le 0$ vs $H_1:\theta>0$ is of the form

$$\varphi(\mathbf x)=\begin{cases}1&,\text{ if }x_{(1)}\ge k\,\,\text{ or }\,\,x_{(n)}\ge 1 \\ 0&,\text{ otherwise }\end{cases}$$

, where $k$ can be easily found so that size of $\varphi$ is $\alpha$.

This is exercise 6.6.18 of Mathematical Statistics by Shao (page 456, 2nd edition).

If you can see this page, then maybe you can use this exercise to verify your final results.

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  • $\begingroup$ Thank you, you are wonderful. I am still digesting your answer. $\endgroup$ May 8, 2019 at 15:08
  • $\begingroup$ You should verify the expression for $\Lambda$ first of all and then proceed accordingly. $\endgroup$ May 8, 2019 at 15:15
  • $\begingroup$ I think I am supposed to use a randomised test, right? If $\theta_1 < x_{(1)}$ then I reject $H_0$ with some probability that makes the size of the test $ \alpha $. $\endgroup$ May 8, 2019 at 15:20
  • $\begingroup$ @NoppaweeApichonpongpan No need for randomisation here. $\endgroup$ May 8, 2019 at 21:25
  • $\begingroup$ No randomisation? I thought I understood but now I am back to being confused. The test I got is ---- reject $H_0$ if $X_{(n)} > 1$ (in this case $Y_{(1)} > \theta_1$) ---- accept $H_0$ if $X_{(n)} < 1$ and $X_{(1)} < \theta_1$ ---- if $X_{(n)} < 1$ and $X_{(1)} > \theta_1$ then reject $H_0$ with probability $\frac{\alpha}{(1-\theta_0)^n}$ This test has size $\alpha$ and I think (but is not sure) is most powerful. Is this incorrect? $\endgroup$ May 9, 2019 at 9:30

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