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: Picture the two families of intervals.
The random variable $X$ always lands inside an interval of length $1$ starting at $\theta$, that is $(\theta,\theta+1)$. Under $H_0$, $\theta\le0$, so this interval sits entirely at or below $(0,1)$. Under $H_1$, $\theta>1$, so the interval sits entirely above $1$.

Step 2: Note the hypotheses do not touch.
Because $\theta\le0$ forces $X\le\theta+1\le1$, and $\theta>1$ forces $X>\theta>1$, any observed value bigger than $1$ rules out $H_0$ completely. That already gives a test region $X>1$ with zero chance of a wrong rejection, using none of our allowed 5 percent error budget.

Step 3: Spend the leftover error budget.
A good test should use all the allowed level, not less, since rejecting for more values of $x$, all of which favor a large $\theta$, only helps catch true alternatives. The natural region to add is values of $x$ just under $1$, so we look for a cutoff $k$ and reject when $X>k$.

Step 4: Pin down k from the worst case inside H0.
Among all $\theta\le0$, the chance of wrongly landing above $k$ is largest when the interval $(\theta,\theta+1)$ sits as high as possible, which happens at $\theta=0$, giving $X\sim\text{Uniform}(0,1)$. The chance of exceeding $k$ is then simply the leftover length, $1-k$. Setting the worst case chance to the allowed level,
\[ 1-k=0.05 \]
\[ k=0.95 \]
So we reject $H_0$ whenever $X>0.95$.

Step 5: Work out the power.
Take any $\theta>1$ under $H_1$. Its interval $(\theta,\theta+1)$ lies completely above $1$, and $1>0.95$, so every single value $X$ can take already satisfies $X>0.95$. The test rejects with certainty, so
\[ \beta=P(X>0.95\mid\theta)=1 \]
for every $\theta$ bigger than $1$, so this constant value is the power asked for.

Step 6: Final number.
\[ 225\beta=225\times1=225 \]
\[ \boxed{225\beta=225} \]