Question

Let \( X_1, X_2, \dots, X_n \) be a random sample from a \( U(\theta, 0) \) distribution, where \( \theta<0 \). If \( T_n = \min(X_1, X_2, \dots, X_n) \), then which of the following sequences of estimators is (are) consistent for \( \theta \)?

• A. \( T_n \)
• B. \( T_n - 1 \)
• C. \( T_n + \frac{1}{n} \)
• D. \( T_n - \frac{1}{n^2} \)

Hint:

For an estimator to be consistent, it must converge to the true value of the parameter as the sample size increases. The minimum of i.i.d. random variables is consistent for the parameter it estimates.

Answer 1

The Correct Option is A, C

Step 1: Understanding consistency.
For an estimator \( T_n \) to be consistent for \( \theta \), it must converge in probability to \( \theta \) as \( n \to \infty \). In this case, since \( T_n \) is the minimum of \( n \) i.i.d. random variables, it is known that the minimum of i.i.d. random variables converges to the true parameter value as \( n \to \infty \). Therefore, \( T_n \) is a consistent estimator for \( \theta \).
Step 2: Analyzing the options.
(A) \( T_n \): This is correct. Since the minimum converges to \( \theta \), \( T_n \) is consistent for \( \theta \).
(B) \( T_n - 1 \): This is incorrect. Shifting \( T_n \) by 1 does not preserve consistency, as it would converge to \( \theta - 1 \), not \( \theta \).
(C) \( T_n + \frac{1}{n} \): This is incorrect. Adding \( \frac{1}{n} \) to \( T_n \) would make the estimator asymptotically biased and not consistent.
(D) \( T_n - \frac{1}{n^2} \): This is also incorrect for the same reason as (C). The shift does not preserve consistency.
Step 3: Conclusion.
The correct answer is \((A)\), as \( T_n \) is a consistent estimator for \( \theta \).