Question

Let \(X\) be a single observation from a distribution having the probability density function
\[f_\theta(x)=\begin{cases}1 & \text{if } \theta<x<\theta+1\\0 & \text{otherwise,}\end{cases}\]
where \(\theta\in(-\infty,\infty)\). For testing \(H_0:\theta\le0\) against \(H_1:\theta>1\), let \(\beta\) be the power of the uniformly most powerful test of level \(0.05\). Then \(225\beta\) equals ______ (answer in integer).

Hint:

Hint:
Since \(\theta\le0\) forces \(X<1\) and \(\theta>1\) forces \(X>1\), find the cutoff \(k\) using \(X\sim\text{Uniform}(0,1)\) at \(\theta=0\), then check that the power is the same for every \(\theta>1\).

Answer 1

Correct Answer: 225

Step 1: Understand the setup.
We have a single observation \(X\) from \(f_\theta(x)=1\) for \(\theta<x<\theta+1\), a uniform distribution shifted by \(\theta\). We test
\[ H_0:\theta\le0 \quad \text{against} \quad H_1:\theta>1 \]
The two hypotheses are separated by a gap, the values of \(\theta\) between \(0\) and \(1\) are covered by neither, which is what makes this problem work out cleanly.

Step 2: Compare the supports under the two hypotheses.
If \(\theta\le0\), the support of \(X\) is \((\theta,\theta+1)\) with \(\theta+1\le1\), so \(X<1\) with certainty.
If \(\theta>1\), the support of \(X\) is \((\theta,\theta+1)\) with \(\theta>1\), so \(X>1\) with certainty.
So any \(x\ge1\) can only happen under \(H_1\), never under \(H_0\).

Step 3: Find the most powerful rejection region.
Because observing \(x>1\) is impossible under every \(\theta\le0\), a test that rejects whenever \(X>1\) already has size \(0\), well below the allowed level \(0.05\). So we can push the boundary of the rejection region below \(1\) to use up the remaining significance level and gain more power. Since large values of \(x\) always favor larger \(\theta\), the optimal region takes the form
\[ \text{Reject } H_0 \text{ if } X>k \]
for some constant \(k\), and this same region turns out to be best for every \(\theta_1>1\), so it is uniformly most powerful for the composite \(H_1:\theta>1\).

Step 4: Fix k using the size condition.
The type I error is largest at \(\theta=0\), the boundary point of \(H_0\), because as \(\theta\) increases toward \(0\) the support \((\theta,\theta+1)\) shifts up and overlaps more with \((k,\infty)\). At \(\theta=0\), \(X\sim\text{Uniform}(0,1)\), so
\[ P(X>k\mid\theta=0)=1-k \]
Setting this equal to the level \(0.05\):
\[ 1-k=0.05 \quad\Rightarrow\quad k=0.95 \]
So the UMP level \(0.05\) test rejects \(H_0\) when \(X>0.95\).

Step 5: Compute the power at any theta greater than 1.
For \(\theta>1\), the support of \(X\) is \((\theta,\theta+1)\), and since \(\theta>1>0.95\), every possible value of \(X\) already exceeds \(0.95\). So
\[ \beta=P(X>0.95\mid\theta)=1 \quad \text{for every } \theta>1 \]
The power is the same, equal to \(1\), for every \(\theta\) in \(H_1\), which is why the question can ask for "the" power \(\beta\) without naming a specific \(\theta\).

Final Answer:
\[ 225\beta=225(1)=225 \] \[ \boxed{225\beta=225} \]