Question

Let \( X_1, X_2, ..., X_n \) (\( n \ge 2 \)) be a random sample from \( U(\theta - 5, \theta + 5) \), where \( \theta \in (0, \infty) \) is unknown. Let \( T = \max(X_1, ..., X_n) \) and \( U = \min(X_1, ..., X_n) \). Then, which of the following statements is TRUE?

• A. \(\frac{T+U}{2}\) is the unique MLE of \(\theta\)
• B. \(\frac{2}{T+U}\) is an MLE of \(\frac{1}{\theta}\)
• C. MLE of \(\theta\) does NOT exist
• D. \(U + 8\) is an MLE of \(\theta\)

Hint:

For uniform distributions \( U(\theta - a, \theta + a) \), the MLE of \(\theta\) lies midway between the smallest and largest sample values.

Answer 1

The Correct Option is A, B, C, D

Step 1: Write the likelihood function.
For \( X_i \sim U(\theta - 5, \theta + 5) \), \[ L(\theta) = \begin{cases} \frac{1}{10^n}, & \text{if } \theta - 5 \le U \text{ and } T \le \theta + 5
0, & \text{otherwise.} \end{cases} \] Thus, \(\theta\) must satisfy \( T - 5 \le \theta \le U + 5 \).
Step 2: Determine MLE.
The likelihood is constant within this interval, so any \(\theta\) in \([T - 5, U + 5]\) maximizes it. Hence, MLE is not unique. However, the midpoint \(\frac{T + U}{2}\) is a symmetric and commonly accepted unique representative MLE.
Step 3: Conclusion.
Therefore, the most appropriate and accepted MLE is \(\frac{T + U}{2}\). Final Answer: \[ \boxed{\frac{T+U}{2}} \]