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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 $1$?

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Your likelihood ratio is

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

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

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

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.

enter image description here

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  • $\begingroup$ 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. $\endgroup$
    – maths54321
    Jan 22, 2021 at 17:13
  • $\begingroup$ @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 $\endgroup$
    – tommik
    Jan 22, 2021 at 17:52
  • $\begingroup$ @maths54321 : I added some details in my answer $\endgroup$
    – tommik
    Jan 23, 2021 at 5:29

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