Solution
A Likelihood Ratio Test for the Endpoint
Show the problem again
Let $X_1,X_2,X_3$ be independent uniform on $[0,\theta]$. To test $H_0:\theta=1$ against $H_1:\theta\ne1$, use the likelihood ratio statistic $\Lambda=\dfrac{\sup_{\theta=1}L(\theta)}{\sup_{\theta>0}L(\theta)}$ and reject when $\Lambda\le 1/8$. Find the probability that this test rejects when the true parameter is $\theta=3/2$. Give a reduced fraction.
Worked solution
With $M=\max X_i$, $L(\theta)=\theta^{-3}\mathbf 1\{\theta\ge M\}$, maximised at $\hat\theta=M$ with value $M^{-3}$. Hence $\Lambda=M^3$ when $M\le1$, and $\Lambda=0$ when $M>1$ (the null likelihood vanishes). So $\Lambda\le1/8$ iff $M\le1/2$ or $M>1$; the rejection region has two pieces (under $H_0$ its size is exactly $1/8$). Under $\theta=3/2$, $P(M\le t)=(t/\theta)^3$: $P(M\le1/2)=(1/3)^3=1/27$ and $P(M>1)=1-(2/3)^3=19/27$. Power $=20/27$.
Source: Original problem written for AxiomIQ. Illustrates a non-regular case where Wilks' theorem does not apply.