How to construct likelihood ratio tests when the null hypothesis is of this form?

Question

For example, I know how to get the likelihood ratio when $H_o: \theta = 3$ against $H_1: \theta > 3$. We derive some rule involving $x$ and a constant from the likelihood ratio, and then find the probability of type I error. But when the null hypothesis is of the form $H_o: \theta \leq 3$, how do I do this? Assume that $X$ is from an exponential distribution with mean $\frac{1}{\theta}$ and I have to construct a LRT with $\alpha=0.05$.

Answer 1

$H_0=3$ and $H_0\leq 3$ are equivalent. This because of the definition of $\alpha$

$$\alpha= \sup_{\theta \in \Theta_0} \pi(\theta)$$

where $\pi(\theta)$ is the power function

thus if you know how to verify this system

$$\begin{cases} H_0: \theta=3\\ H_1: \theta>3 \end{cases}$$

verify this one in the same way.

$$\begin{cases} H_0: \theta\leq3\\ H_1: \theta>3 \end{cases}$$

in an informal way: The null hypotesis is always considered TRUE until you have enough data to reject it...thus you assume as $H_0$ the best value you can choose...so if $H_0$ is $\theta \leq 3$ you choose $\theta=3$ as the best value to be verified

If your model belongs to the exponential family (as it is the example you posted) this test is well explained here, on page 422, theorem 5

If you read next page, 423 the example 13 is exactly yours! What the book does not state is that the resulting integral can be solved via chi-squared distribution, but I think you perfectly know that.