Question

Let \( X_1, X_2 \) be a random sample from a distribution having probability density function 

• A. \( \phi_1(X_1, X_2) = \begin{cases} \frac{X_1 + X_2 - 1}{X_1 + X_2}, & \text{if } X_1 + X_2 > 1, \\ 0, & \text{otherwise}. \end{cases} \)
• B. \( \phi_2(X_1, X_2) = \begin{cases} \frac{2X_1 + 2X_2 - 1}{2(X_1 + X_2)}, & \text{if } X_1 + X_2 > 1, \\ 0, & \text{otherwise}. \end{cases} \)
• C. \( \phi_3(X_1, X_2) = \begin{cases} \frac{3X_1 + 3X_2 - 1}{3(X_1 + X_2)}, & \text{if } X_1 + X_2 > 1, \\ 0, & \text{otherwise}. \end{cases} \)
• D. \( \phi_4(X_1, X_2) = \begin{cases} \frac{4X_1 + 4X_2 - 1}{4(X_1 + X_2)}, & \text{if } X_1 + X_2 >1, \\ 0, & \text{otherwise}. \end{cases} \)

Hint:

When analyzing tests with different forms, focus on how the power function is derived and check if the form of the test aligns with the original test’s power function.

Answer 1

The Correct Option is A

Step 1: Understand the original test \( \phi(X_1, X_2) \).
The test \( \phi(X_1, X_2) \) is defined as: \[ \phi(X_1, X_2) = \begin{cases} 1, & \text{if } X_1 > 1, \\ 0, & \text{otherwise}. \end{cases} \] This test is checking whether \( X_1 \) is greater than 1, and it essentially ignores the value of \( X_2 \).
Step 2: Understand the power function of \( \phi(X_1, X_2) \).
The power function of a test gives the probability of rejecting the null hypothesis \( H_0 \) (i.e., the test statistic is greater than a critical value) for various values of the parameter \( \theta \).
For \( \theta = 1 \) (the null hypothesis), the probability of rejecting \( H_0 \) is: \[ P(X_1 > 1 \mid \theta = 1) = \int_1^\infty \frac{1}{1} e^{-x/1} \, dx = e^{-1}. \] For \( \theta > 1 \), the probability of rejecting \( H_0 \) is: \[ P(X_1 > 1 \mid \theta) = 1 - F_X(1) = 1 - (1 - e^{-1/\theta}) = e^{-1/\theta}. \]
Step 3: Analyze the options.
Option (A): This test depends on \( X_1 + X_2 \), and it has the same form as the original test because the sum of \( X_1 \) and \( X_2 \) will determine the power function in a way that matches the original test for large \( \theta \). Hence, option (A) is correct.
Option (B), (C), and (D): These tests differ in the factors used in the formula, which would result in different power functions that do not match the power function of the original test \( \phi \).
Thus, the correct answer is \( \boxed{\text{(A)}} \).