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I found a couple of questions on this site (Most powerful test of simple vs. simple in $\mathrm{Unif}[0, \theta]$ and UMP for $U(0,\theta)$ (simple x simple hypothesis)) that are similar to my problem, but I don't understand enough about hypothesis testing to translate the discussion in those links to my situation.

Problem:

Let $Y_1, \dots, Y_n$ be a random sample from a uniform distribution on the interval $[0, \theta]$ where $\theta > 0$ is unknown. Find the most powerful test for the null hypothesis $H_0 \colon \theta = 1$ against the alternative hypothesis $H_1 \colon \theta = 4$ which never rejects a true null hypothesis. Find the power of this most powerful test when $n = 4$.

My attempt:

I initially considered the ratio of likelihoods $\dfrac{L_{\boldsymbol{Y}}(\theta_1 ; \boldsymbol{Y})}{L_{\boldsymbol{Y}}(\theta_0 ; \boldsymbol{Y})}$ but this didn't seem to lead anywhere. Then I decided to use the following test, based purely on intuition: reject $H_0$ if and only if the sample maximum $Y_{(n)} > 1$.

It is my understanding that this test is actually uniformly most powerful for the composite alternative $H_1^{\prime} \colon \theta > 1$ because the rejection region does not depend on the value of $\theta_1$ ($4$ in this case). Therefore the test must certainly be most powerful for the null and alternative hypotheses as stated in the problem.

I am pretty sure the test will never reject a true null hypothesis (i.e. has size zero) because, well, by construction it will only reject $H_0$ when the sample maximum is greater than $1$, which means that $H_0$ must be false.

As for the power of the test, I reason as follows (note that according to my course notes, the phrase "power of the test" refers to the value of the function $\beta(\theta) = P( \text{reject } H_0)$ in the situation that $H_1$ is true):

\begin{align} \text{Power of the test} &= P( \text{reject } H_0)\\ &= P( Y_{(n)} > 1)\\ &= 1 - F_{Y_{(n)}}(1)\\ &= 1 - \left( \frac{1}{4} \right)^{\!\!4} \text{ (plugging in d.f. of } Y_{(n)} \text{ when } n = 4 \text{ and } H_1 \text{ is true)}\\ &= \frac{255}{256}. \end{align}

Question:

Is this solution correct?

Thanks.

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  • $\begingroup$ For a large sample size, it probably doesn't matter, but $Y_{(n)}$ is a biased estimator of $\theta.$ $\hat\theta=(n+1)Y_{(n)}/n$ is unbiased, so it might be better to use $\hat\theta$ instead, particularly with lower sample sizes. $\endgroup$ Aug 18, 2021 at 17:35

1 Answer 1

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Yes, it is correct. You can derive this test from the likelihood ratio. The likelihood $L_\theta$ is $\theta^n$ if all $Y_i\leq\theta$ and 0 otherwise, so the likelihood ratio is $(1/4)^n$ if all $y_i\leq 1$ and $0$ otherwise.

The most powerful test must choose $\theta=1$ if $Y_{(n)}\leq 1$ and $\theta=4$ otherwise, since those correspond to the only possible values of the likelihood ratio and as you show, this test has size zero and power 255/256.

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  • $\begingroup$ What would the level of this test be? $\endgroup$ Aug 18, 2021 at 17:26
  • $\begingroup$ Zero. The test never rejects $\theta=1$ when it's true $\endgroup$ Aug 18, 2021 at 23:38

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