Question

Let $X_1, X_2, \ldots, X_n$ be a random sample from a distribution with probability density function \[ f_{\theta}(x) = \begin{cases} \theta (1 - x)^{\theta - 1}, & 0 < x < 1, \\ 0, & \text{otherwise}, \end{cases} \theta > 0. \] To test $H_0: \theta = 1$ against $H_1: \theta > 1$, the uniformly most powerful (UMP) test of size $\alpha$ would reject $H_0$ if 
 

• A. $-\sum_{i=1}^n \log_e (1 - X_i)^2 < \chi^2_{2n, 1 - \alpha}$
• B. $-\sum_{i=1}^n \log_e (1 - X_i)^2 < \chi^2_{n, 1 - \alpha}$
• C. $-\sum_{i=1}^n \log_e (1 - X_i)^2 < \chi^2_{2n, \alpha}$
• D. $\sum_{i=1}^n \log_e (1 - X_i)^2 < \chi^2_{n, \alpha}$

Hint:

For one-parameter exponential family distributions, transformations of the likelihood ratio often follow chi-square laws, making it easier to construct UMP tests.

Answer 1

The Correct Option is A

Step 1: Identify the distribution.
The given pdf is \[ f_{\theta}(x) = \theta (1 - x)^{\theta - 1}, 0 < x < 1. \] This is the pdf of a Beta(1, $\theta$) distribution.

Step 2: Apply transformation.
Let $Y_i = -2 \log(1 - X_i)$. Then, under $\theta = 1$, the variable $Y_i$ follows an Exponential(1) distribution with mean $2$. Hence, the sum \[ T = -2 \sum_{i=1}^{n} \log(1 - X_i) \] follows a Chi-square distribution with $2n$ degrees of freedom under $H_0$.

Step 3: Derive the critical region.
For the alternative $H_1: \theta > 1$, larger values of $\theta$ make $(1 - X_i)$ smaller on average, thus increasing $T$. Hence, we reject $H_0$ for large values of $T$. The critical region is: \[ T > \chi^2_{2n, 1 - \alpha}. \]

Step 4: Express in terms of $X_i$.
Since $T = -2 \sum_{i=1}^{n} \log(1 - X_i)$, \[ \boxed{-\sum_{i=1}^{n} \log_e (1 - X_i)^2 < \chi^2_{2n, 1 - \alpha}.} \]

Step 5: Conclusion.
The UMP test rejects $H_0$ for large values of $-\sum \log(1 - X_i)^2$, i.e., as given in option (A).