# Critical region for an uniform distribution

Let $$X_1, X_2, ... , X_n$$ be a random sample from the uniform distribution over $$[0, \theta]$$. Suppose we wish to test $$H_0 : \theta = 5$$ versus $$H_A : \theta < 5$$ at significance level $$\alpha = 0.05$$. Use the LRT to find the critical region.

Here is my attempt:

\begin{align*} T = \frac{L(\theta_0 = 5)}{L(\theta_A < 5)} & = \frac{\Pi_{i = 1}^n \frac{1}{b-a}}{\Pi_{i = 1}^n \frac{1}{b-a}} \\ & = \frac{\frac{1}{(5-a)^n}}{\frac{1}{(\theta_A-a)^n}} \\ & = \frac{(\theta_A-a)^n}{(5-a)^n} \end{align*}

However, I am not sure how to proceed next. I know that we reject the null hypothesis if $$T < c$$ where $$c$$ is some constant. In my book they give an example where they now take $$\ln$$ on both sides and if I do so I get $$\ln(T) = - n \ln(\theta_A - a) + n \ln(5 - a)) < \ln(c)$$

but what am I supposed to solve for?

• Is this for a homework assignment? Mar 23, 2021 at 15:55
• No, it is just homework that I have to prepare for my class on Thursday (we go through all of the exercises then but I just want to understand everything before class) Mar 23, 2021 at 15:57

The likelihood ratio is $$\lambda(x) = \frac{\sup_{\theta\in\Theta_0} L(\theta\mid x)}{\sup_{\theta\in\Theta} L(\theta\mid x)}.$$ In this case $$\Theta_0 = \{5\}$$, $$\Theta = [0,5]$$, and your likelihood is $$L(\theta\mid x) = \theta^{-n}\mathbf 1_{x_\max \leq \theta }.$$ On $$[0,5]$$ $$\theta^{-n}$$ is strictly decreasing so we want the smallest $$\theta$$ possible such that we don't violate the support, and this leads to $$\sup_{\theta\in\Theta} L(\theta\mid x) = L(x_\max \mid x)$$. This means our likelihood ratio is $$\lambda(x) = \frac{L(5 \mid x)}{L(x_\max \mid x)} = \left(\frac{5}{x_\max}\right)^{-n}.$$ $$x_\max$$ is a sufficient statistic for $$\theta$$ here so it's nice to see that our LRT only depends on this sufficient statistic. You got this part already but I think your notation was a little unclear since $$a$$ and $$b$$ weren't really defined.
For the rejection region, we can get there by thinking about what we're trying to do. We'll reject the null hypothesis if $$\lambda(x)$$ is small, because this means there's a much higher likelihood outside of $$\Theta_0$$ than in it. To get our threshold then we know that we want the probability of a rejection under $$H_0$$ to be $$\alpha$$, so this means we need to solve $$P(\lambda(X) \leq c \mid \theta \in \Theta_0) = \alpha$$ for $$c$$. The null hypothesis is simple here so things are a little easier: $$P\left(\lambda(X) \leq c \mid \theta \in \Theta_0\right) = P\left(\left(\frac{5}{X_\max}\right)^{-n} \leq c\right) = \alpha.$$ At this point can you finish the problem?
• where from the question stem do you find $\Theta = [0,5]$ ? Jul 10, 2022 at 16:48