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Suppose $X_1,X_2,...,X_n$ is a random sample from $U(0,\theta)$. We need to construct a likelihood ratio size $\alpha$ test to test $H_0:\theta=\theta_0$ against $H_{1}: \theta \neq \theta_{0}$

My approach:

$\sup_{\theta}\frac{1}{\theta^n}= \frac{1}{(X_{(n)})^n}$ Thus the likelihood ratio is : $\lambda=(\frac{X_{(n)}}{\theta_0})^n$ Now if $X_{(n)}<\theta_0$ , then $\lambda$ is a decreasing function of $X_{(n)}$.So the LRT size $\alpha$ test criterion $\lambda<c $ is then equivalent to $X_{(n)} >k$ and if $X_{(n)}>\theta_0$ , then $\lambda$ is an increasing function of $X_{(n)}$ and the LRT criterion yields $ X_{(n)}<k'$ Now is this correct? How will I find $k,k'$ for a fixed size $\alpha$?

StubbornAtom
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  • Certainly $X_{(n)}>\theta_0$ rejects the null immediately, and only $X_{(n)}<\theta_0$ case is of interest; start from thinking what is the distribution of $X_{(n)}$ under the null. – kludg May 07 '19 at 03:43
  • Yes, I found the value of $k$ ,it is coming to be $\theta_0(\alpha)^{\frac{1}{n}}$. But then the both sided test criteria matches with the one sided test in order to test $H_1: \theta>\theta_0$ –  May 07 '19 at 04:10
  • Is there a problem ? –  May 07 '19 at 04:11
  • Since you are asking about LR test probably this is asymptotic test, and LR statistic asymptotically is chi square with 1 degree of freedom; LR test is not the best for the problem since it is naturally 2-sided, and I guess it can be asymptotic only, so the test threshold value should be taken from chi square tables or better from software like R. – kludg May 07 '19 at 06:18

1 Answers1

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The likelihood ratio criterion is more precisely given by

$$\lambda(x_1,\ldots,x_n)=\begin{cases} \left(\frac{x_{(n)}}{\theta_0}\right)^n &,\text{ if } x_{(n)}\le \theta_0 \\ 0 &,\text{ if } x_{(n)}>\theta_0 \end{cases}$$

Clearly $\lambda$ is non-decreasing in $x_{(n)}$ for $0<x_{(n)}\le\theta_0$. Since we reject the LR test for small values of $\lambda$, the critical region has the following form for some $k\le\theta_0$: $$\left\{(x_1,\ldots,x_n)\mid x_{(n)}>\theta_0\,\text{ or }\,x_{(n)}<k\right\}$$

Imposing a size $\alpha$ restriction on the test gives $k=\theta_0 \alpha^{1/n}$. Not surprisingly, the LR test coincides with the UMP test for testing the same hypothesis.

StubbornAtom
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