# Determining the significance level for a Neyman-Pearson hypothesis test when $X_1,...,X_n \sim \text{Uniform}(0,\theta)$?

We want to test $$H_0: \theta=\theta_0$$ against $$H_1: \theta=\theta_1$$ where $$0<\theta_1<\theta_0$$ and where $$X_1,...,X_n\sim$$ Uniform$$(0,\theta)$$ are IID. We have critical value $$c$$.

My problem is: I want to justify why the significance level, $$\alpha$$, must be such that $$\alpha\geq\frac{\theta_1^n}{\theta_0^n}$$ in the case where $$c\leq\frac{\theta_0^n}{\theta_1^n}$$

I've computed the Neyman-Pearson test statistic to be $$T(\textbf{x})=\frac{f(\textbf{x};\theta_1)}{f(\textbf{x};\theta_0)}=\frac{\frac{1}{\theta_1^n}\mathbb{1}\{x_1,...,x_n\in[0,\theta_1]\}}{\frac{1}{\theta_0^n}\mathbb{1}\{x_1,...,x_n\in[0,\theta_0]\}}= \frac{\theta_0^n \mathbb{1}\{x_1,...,x_n\in[0,\theta_1]\}}{\theta_1^n\mathbb{1}\{x_1,...,x_n\in[0,\theta_0]\}}$$

This implies that the power function is $$\mathbb{P}\bigg( \frac{\theta_0^n \mathbb{1}\{x_1,...,x_n\in[0,\theta_1]\}}{\theta_1^n\mathbb{1}\{x_1,...,x_n\in[0,\theta_0]\}}\geq c ;\theta\bigg)$$

Now in the case where $$x_1,...,x_n\in[0,\theta_1]$$, if $$\frac{\theta_0^n}{\theta_1^n}\geq c$$ then clearly the power function is $$1$$, and if $$\frac{\theta_0^n}{\theta_1^n}< c$$ then clearly it is $$0$$.

Now in the case where there exists $$i$$ such that $$x_i\in(\theta_1,\theta_0]$$ then the power function becomes $$\mathbb{P}(0\geq c)=0$$.

So going back to the problem: when $$c\leq\frac{\theta_0^n}{\theta_1^n}$$, how are we able to deduce from what I've done that $$\alpha=\mathbb{P}(T(\textbf{X})\geq c;\theta_0)\geq \frac{\theta_1^n}{\theta_0^n}$$. Perhaps I've done something wrong because all I am getting is that the power function is either $$0$$ or $$1$$ and hence $$\alpha$$ is either $$0$$ or $$1$$?

$$\frac{L(\theta_1|\mathbf{x})}{L(\theta_0|\mathbf{x})}=\dots=\begin{cases} \Big(\frac{\theta_0}{\theta_1}\Big)^n, & \text{if 0

thus the likelihood ratio is monotonic and we can apply a well known theorem and the critical region is

$$C=\{\mathbf{x};x_{(n)}

Thus

$$\alpha=\int_0^c \frac{n y^{n-1}}{\theta_0^n}dy=\frac{c^n}{\theta_0^n}$$

Now I think it is self evident that you cannot justify your statement.

Counterexample:

Set $$\theta_0=3$$,$$\theta_1=2$$, $$n=2$$

If $$c<\Big(\frac{3}{2}\Big)^2=\frac{9}{4}$$, say $$c=\frac{6}{4}$$, $$\alpha=\frac{1.5^2}{9}=\frac{1}{4}<\frac{4}{9}$$

This is a graphical explanation: The two drawing represent the one sided test with your uniform. I type error (significance level) is the purple area. As you can see,

• in the left drawing, that is the case when $$\theta_1<\theta_0$$ as $$c$$ decreases, so do $$\alpha$$.

• in the right drawing, that is the case when $$\theta_1>\theta_0$$ as $$c$$ decreases, $$\alpha$$ increases.

Between $$c$$ and $$alpha$$ there is an easy relation:

$$\alpha=\frac{c^n}{\theta_0^n}$$

$$\alpha=1-\frac{c^n}{\theta_0^n}$$

respectively, in the two cases. • So my statement that $\alpha\geq \frac{\theta_1^n}{\theta_0^n}$ is not true? Do you have any hints on how to determine $\alpha$ in both the cases $c\leq\frac{\theta_0^n}{\theta_1^n}$ and $c>\frac{\theta_0^n}{\theta_1^n}$? Thanks. Jan 22, 2021 at 17:13
• @maths54321 if you read well my answer $\alpha=\int_0^c f_T(t)dt$ thus if $c$ decreases also $\alpha$ decreases. Different situation will be if $\theta_1>\theta_0$ but this is not your case Jan 22, 2021 at 17:52
• @maths54321 : I added some details in my answer Jan 23, 2021 at 5:29