Question

Let \(X_1, X_2, ..., X_n\) be a random sample from an exponential distribution with probability density function \[ f(x; \theta) = \begin{cases} \theta e^{-\theta x}, & x>0,
0, & \text{otherwise,} \end{cases} \] where \(\theta \in (0, \infty)\) is unknown. Let \(\alpha \in (0,1)\) be fixed and let \(\beta\) be the power of the most powerful test of size \(\alpha\) for testing \(H_0: \theta = 1\) against \(H_1: \theta = 2\). Consider the critical region \[ R = \left\{ (x_1, x_2, ..., x_n) \in \mathbb{R}^n : \sum_{i=1}^n x_i>\frac{1}{2}\chi^2_{2n}(1-\alpha) \right\}, \] where for any \(\gamma \in (0,1)\), \(\chi^2_{2n}(\gamma)\) is a fixed point such that \( P(\chi^2_{2n}>\chi^2_{2n}(\gamma)) = \gamma. \) Then, the critical region \(R\) corresponds to the

• A. most powerful test of size \(\alpha\) for testing \(H_0: \theta = 1\) against \(H_1: \theta = 2\)
• B. most powerful test of size \(1 - \alpha\) for testing \(H_0: \theta = 2\) against \(H_1: \theta = 1\)
• C. most powerful test of size \(\beta\) for testing \(H_0: \theta = 2\) against \(H_1: \theta = 1\)
• D. most powerful test of size \(1 - \beta\) for testing \(H_0: \theta = 1\) against \(H_1: \theta = 2\)

Hint:

For exponential families, the Neyman–Pearson lemma gives a rejection region based on the sum of observations, often expressed through chi-square quantiles.

Answer 1

The Correct Option is D

Step 1: Write the likelihood ratio.
For exponential distribution \( f(x; \theta) = \theta e^{-\theta x} \), the likelihood function for the sample is \[ L(\theta) = \theta^n e^{-\theta \sum_{i=1}^n x_i}. \] Hence, the likelihood ratio is \[ \Lambda(x_1, ..., x_n) = \frac{L(1)}{L(2)} = \frac{1^n e^{-\sum x_i}}{2^n e^{-2\sum x_i}} = \frac{e^{\sum x_i}}{2^n}. \]
Step 2: Apply Neyman–Pearson lemma.
The most powerful test for testing \( H_0: \theta = 1 \) vs. \( H_1: \theta = 2 \) rejects \(H_0\) for large values of \(\sum x_i\). Therefore, the rejection region has the form \[ \sum_{i=1}^n x_i>c. \]
Step 3: Determine the critical value.
Under \(H_0: \theta = 1\), we have \(2\sum X_i \sim \chi^2_{2n}\). Hence, for size \(\alpha\), \[ P_{H_0}\left( \sum_{i=1}^n X_i>\frac{1}{2}\chi^2_{2n}(1 - \alpha) \right) = \alpha. \] Thus, the given region corresponds exactly to a level-\(\alpha\) test.
Step 4: Identify the test type.
The region rejects \(H_0\) when \(\sum X_i\) is large, appropriate for \(H_1: \theta = 2\) (larger rate implies smaller means). Hence, \(R\) is the most powerful test of size \(\alpha\) for \(H_0: \theta = 1\) vs. \(H_1: \theta = 2\). Final Answer: \[ \boxed{\text{(A) most powerful test of size } \alpha \text{ for testing } H_0: \theta = 1 \text{ against } H_1: \theta = 2.} \]