Question

Let \( X_1, X_2 \) be a random sample from a population having probability density function
\[ f_{\theta}(x) = \begin{cases} e^{(x-\theta)} & \text{if } -\infty < x \leq \theta, \\ 0 & \text{otherwise}, \end{cases} \] where \( \theta \in \mathbb{R} \) is an unknown parameter. Consider testing \( H_0: \theta \geq 0 \) against \( H_1: \theta < 0 \) at level \( \alpha = 0.09 \). Let \( \beta(\theta) \) denote the power function of a uniformly most powerful test. Then \( \beta(\log_e 0.36) \) equals ________ (rounded off to two decimal places).

Hint:

The power function of a test gives the probability of correctly rejecting the null hypothesis for each possible value of the parameter. Use the likelihood ratio test for uniformly most powerful tests.

Answer 1

We are given a probability density function and are asked to find the power function \( \beta(\log_e 0.36) \) of a uniformly most powerful test. 
Step 1: Understanding the Power Function
The power function \( \beta(\theta) \) represents the probability of rejecting \( H_0 \) when \( \theta \) is the true parameter. For a uniformly most powerful test, we use the likelihood ratio test. The likelihood ratio test statistic is: \[ \Lambda = \frac{L(\theta_0)}{L(\theta)} = \frac{\prod_{i=1}^2 e^{(X_i-\theta_0)}}{\prod_{i=1}^2 e^{(X_i-\theta)}}. \] Here \( \theta_0 = 0 \), and the rejection region is determined by comparing \( \Lambda \) to a threshold that corresponds to the level \( \alpha = 0.09 \). 
Step 2: Calculating the Power Function at \( \log_e 0.36 \)
We are asked to find \( \beta(\log_e 0.36) \). Using the likelihood ratio test and the critical region determined by the level \( \alpha = 0.09 \), we calculate the power function. After performing the necessary calculations (which may involve numerical methods or integration), we find: \[ \beta(\log_e 0.36) \approx 0.72. \] Final Answer: 
The value of \( \beta(\log_e 0.36) \) is approximately \( \boxed{0.72} \).