I'm having trouble defining the reject region based on the generalized likelihood ratio test. This is from a question of past exam I'm self-studying and still have the doubt, given that I got it wrong.
Let $X_{1},\ldots,X_{n}$ a sample from an exponential distribution $X \sim \text{Exp}(\theta)$ with $\theta = \mathbb{E}[X]$. The parameter space is $\Omega = \{\theta \mid\theta>0\}.$
I need to contrast the next hypothesis: $H_{0}:\theta \geq \theta_{0}$ vs $H_{1}: \theta< \theta_{0}$.
The set of the null hypothesis is: $\Omega_{0} = \{\theta \mid\theta \geq \theta_{0}\}$ and the set of the alternative hypothesis is: $\Omega_{1} = \{\theta \mid\theta < \theta_{0}\}$. So that $\Omega = \Omega_{0} \cup\Omega_{1}.$
The likelihood function for $\theta$ is given by: $$ L(\theta) = \theta^{-n} \exp\{ -\theta^{-1} \bar{X} \} $$
I know the generalized likelihood ratio is given by $$\Lambda = \frac{\sup_{\theta \in \Omega_{0}} L (\theta)}{\sup_{\theta \in \Omega} L (\theta)}$$
and serves as a test statistic. For the denominator, we have that the MLE of $\theta$ is given by $\hat{\theta} = \bar{X}$, and with this the denominator is $$L(\hat{\theta}) = \bar{X}^{-n}\exp\{ -\bar{X}^{-1}n\bar{X} \} = \bar{X}^{-n}\exp\{ -n \}$$
My problem is: how do I solve for the numerator $\sup_{\theta \in \Omega_{0}} L (\theta)$ ? Because I already know how to solve the test given a simple null hypothesis $H_{0}: \theta = \theta_{0}$ and I'm really stucked with this problem.
Thank you for your help.
Best Answer
Unrestricted MLE of $\theta$ is as you say $\hat\theta=\overline X$, the sample mean.
Now under the restriction $\theta\ge\theta_0$, argue that MLE of $\theta$ must be $$\hat{\hat\theta}=\begin{cases}\hat\theta&,\text{ if }\hat\theta\ge\theta_0 \\ \theta_0&,\text{ if }\hat\theta<\theta_0\end{cases}$$
So depending upon whether $\overline X\ge \theta_0$ or $\overline X<\theta_0$, the likelihood ratio statistic takes the form
\begin{align} \Lambda=\frac{\sup_{\theta\ge\theta_0} L(\theta)}{\sup_{\theta}L(\theta)}&=\frac{L(\hat{\hat\theta})}{L(\hat\theta)} \\&=\begin{cases}1&,\text{ if }\hat\theta\ge\theta_0 \\\\ \frac{L(\theta_0)}{L(\hat\theta)}&,\text{ if }\hat\theta<\theta_0\end{cases} \end{align}
Now it is a matter of studying this ratio as a function of $\overline X$ when $\hat\theta<\theta_0$. Remember to reject $H_0$ for small values of $\Lambda$. The case corresponding to $\hat\theta\ge\theta_0$ leads to trivial acceptance of $H_0$.