Question

Let \( X_1, X_2, \dots, X_n \ (n \ge 2) \) be a random sample from a distribution with probability density function \[ f(x; \theta) = \begin{cases} \dfrac{1}{2\theta}, & -\theta \le x \le \theta, \\ 0, & |x| > \theta, \end{cases} \] where \( \theta \in (0, \infty) \) is unknown. If \( R = \min\{X_1, X_2, \dots, X_n\} \) and \( S = \max\{X_1, X_2, \dots, X_n\} \), then which of the following statements is/are TRUE?

• A. \( (R, S) \) is jointly sufficient for \( \theta \)
• B. \( S \) is an MLE of \( \theta \)
• C. \( \max\{|X_1|, |X_2|, ..., |X_n|\} \) is a complete and sufficient statistic for \( \theta \)
• D. Distribution of \( \dfrac{R}{S} \) does NOT depend on \( \theta \)

Hint:

For uniform families over symmetric intervals, the MLE and sufficient statistic are typically the extreme (maximum absolute) sample values.

Answer 1

The Correct Option is A, C, D

Step 1: Understanding the model.
The given distribution is uniform over the symmetric interval \([- \theta, \theta]\). Hence, the joint pdf is: \[ L(\theta; x_1, \dots, x_n) = \begin{cases} (2\theta)^{-n}, & \text{if } -\theta \le x_i \le \theta \ \forall i, \\ 0, & \text{otherwise.} \end{cases} \]
Step 2: Finding the MLE.
For the likelihood to be non-zero, we need \(\theta \ge \max_i |x_i|\). Since \(L\) is decreasing in \(\theta\), the MLE is \[ \hat{\theta} = \max_i |x_i|. \] Thus, option (B) is TRUE.
Step 3: Sufficiency and completeness.
The likelihood depends on the sample only through \(\max_i |x_i|\), so it is a sufficient statistic. For the uniform family of this type, this statistic is also complete. Hence, option (C) is TRUE.
Step 4: Distributional independence.
Since both \(R\) and \(S\) are scaled by \(\theta\) (i.e., \(R/\theta, S/\theta\) have distributions independent of \(\theta\)), the ratio \(R/S\) also does not depend on \(\theta\). Therefore, option (D) is TRUE.
Step 5: Analyze (A).
\((R, S)\) is not minimal sufficient because the joint pdf depends only on \(\max |X_i|\), not both endpoints separately. Thus, (A) is FALSE. Final Answer: \[ \boxed{(B), (C), (D)} \]