Finding the UMP size $0.1$ test with $U\sim\text{Unif}(\theta, \theta+1)$

$$U$$ is a uniform $$(\theta, \theta+1)$$ random variable, where where $$\theta\in(−0.5,0.5)$$. Consider testing

$$H_0:\theta\geq0$$

$$H_1: \theta \lt0$$

(a) Give the test function of a UMP size $$0.1$$ test

(b) Give the power of a UMP size $$0.1$$ test against the specific alternative of $$\theta=−0.3$$

(c) For the test requested in part (a), give the p-value if the observed value of $$U$$ is $$−0.1$$

My try:

(a) We have that $$T(\vec{U})=U_{(n)}=U$$ is a univariate sufficient statistic. The cdf of $$T$$ is

$$F_T(t\mid\theta)= \begin{cases} 1 & t \geq \theta+1 \\ t-\theta & \theta \lt t \lt \theta+1 \\ 0 & t\leq \theta \end{cases}$$

and the pdf of $$T$$ is $$f_T(t\mid\theta)=I_{(\theta,\theta+1)}(t)$$

If $$\theta_1\lt\theta_2$$ we have

$${\{t:f_T(t\mid\theta_1)\gt0 \text{ or } f_T(t\mid\theta_2)\gt0}\}=(\theta_1,\theta_1+1)\cup(\theta_2,\theta_2+1)$$

For $$\theta_1\lt\theta_2$$ and $$t\in(\theta_1,\theta_1+1)\cup(\theta_2,\theta_2+1)$$ we have

$$\frac{f_T(t\mid\theta_2)}{f_T(t\mid\theta_1)}=\frac{I_{(\theta_2,\theta_2+1)}(t)}{I_{(\theta_1,\theta_1+1)}(t)}= \begin{cases} \infty & t\in[\theta_1+1,\theta_2+1) \\ 1 & t\in(\theta_2,\theta_1+1) \\ 0 & t\in(\theta_1,\theta_2] \end{cases}$$

and so the family of pdfs of $$T$$ has a nondecreasing, monotone likelihood ratio. Hence, the test which rejects $$H_0$$ iff $$T(\vec{U})=U_{(n)}=U\leq c$$ is a UMP size $$0.1$$ test, where

$$0.1=\mathsf P_{\theta=0}(T\leq c)= \begin{cases} 1 & c \geq 0 \\ c & 0 \lt c \lt 1 \\ 0 & c\leq 0 \end{cases}$$

So for a size $$0.1$$ UMP test we need $$c=0.1$$. Hence we reject iff $$T=U$$ assumes a value in $$(-0.5,0.1]$$. Hence the test function of a UMP size $$0.1$$ test is

$$\phi(t)= \begin{cases} 1 & t\in(-0.5,0.1] \\ 0 & t \in(0.1,1.5) \end{cases}$$

Is this a valid solution thus far?

(b)

We have

$$\text{Power} = \mathsf P(\text{reject } H_0\mid H_1 \text{ true})=\mathsf P(U\in(-0.5,0.1]\mid\theta=-0.3)=0.4$$

(c) We have

$$\mathsf P(\text{reject } H_0\mid H_0 \text{ true})=\mathsf P(U\in(-0.5,0.1]\mid\theta=0)=0.1$$

Am I correct in assuming that $$\theta=0$$ even though under the null hypothesis $$\theta\geq0$$?

Edit:

I think for (c) I should actually be finding the probability of observing something as extreme or more extreme than $$U=-0.1$$ under the null hypothesis. But since under the null, $$U\sim\text{Unif}(0,1)$$, we cannot observe $$U=-0.1$$ so the p-value would be $$0$$? Is my reasoning correct?

• Since about 80% of this post is neither a question nor relevant to your only question (which concerns part (b)), could you edit it down to focus on the actual question? – whuber Nov 28 '18 at 19:18
• Well (a) was necessary to do part (b) and I think I did (a) right so I don't see how removing it would be beneficial. Perhaps I'm misunderstanding. – Remy Nov 28 '18 at 19:26
• In your answer to (a), what will be your conclusion if $T$ is non-positive? If it exceeds $0.1$? – whuber Nov 28 '18 at 19:46
• It cannot happen under the null but I would think you'd reject since negative values would be even more extreme than $0$. If it exceeds $0.1$, we'd fail to reject. – Remy Nov 28 '18 at 19:50
• Yes, thank you. I think the mistake I made initially was that the rejection region should only be upper-bounded. – Remy Nov 29 '18 at 15:19