Question

Let \( X \) be a random variable having a probability density function \( f \in \{ f_0, f_1 \} \), where
\[ f_0(x) = \begin{cases} 1, & 0 \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases} \quad \text{and} \quad f_1(x) = \begin{cases} \frac{1}{2}, & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases} \] For testing the null hypothesis \( H_0 : f = f_0 \) against \( H_1 : f = f_1 \), based on a single observation on \( X \), the power of the most powerful test of size \( \alpha = 0.05 \) equals

• A. 0.425
• B. 0.525
• C. 0.625
• D. 0.725

Hint:

The power of a test is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true. It can be calculated by determining the rejection region and evaluating the test statistic under the alternative hypothesis.

Answer 1

The Correct Option is B

Step 1: Set up the likelihood ratio test.
The most powerful test of size \( \alpha = 0.05 \) can be derived using the likelihood ratio test. The test statistic is the likelihood ratio: \[ \Lambda(x) = \frac{f_1(x)}{f_0(x)} \] We reject \( H_0 \) if \( \Lambda(x) \) exceeds a certain threshold.
Step 2: Calculate the critical value.
The critical value is determined using the size of the test \( \alpha = 0.05 \), which gives the threshold for rejecting \( H_0 \).
Step 3: Final calculation.
The power of the test is the probability of rejecting \( H_0 \) when \( f = f_1 \), which is approximately 0.525.