Question

Let \( X_1, X_2, \dots, X_n \ (n \ge 2) \) be a random sample from a distribution with probability density function \[ f(x; \theta) = \begin{cases} \theta x^{\theta - 1}, & 0 \le x \le 1, \\ 0, & \text{otherwise,} \end{cases} \] where \( \theta \in (0, \infty) \) is unknown. Then, which of the following statements is/are TRUE?

• A. Cramer-Rao lower bound, based on \( X_1, X_2, ..., X_n \), for the estimator \( \theta^3 \) is \( \dfrac{9\theta^6}{n} \)
• B. Cramer-Rao lower bound, based on \( X_1, X_2, ..., X_n \), for the estimator \( \theta^3 \) is \( \dfrac{9\theta^4}{n} \)
• C. There does NOT exist any unbiased estimator of \( \dfrac{1}{\theta} \) which attains the Cramer-Rao lower bound
• D. There exists an unbiased estimator of \( \dfrac{1}{\theta} \) which attains the Cramer-Rao lower bound

Hint:

In power-law distributions like \(f(x; \theta) = \theta x^{\theta-1}\), Fisher information for one sample is \(1/\theta^2\). Use \(g'(\theta)\) to find CRLB for any transformation \(g(\theta)\).

Answer 1

The Correct Option is A, D

Step 1: Find the Fisher Information.
For a single observation: \[ \ln f(x; \theta) = \ln \theta + (\theta - 1)\ln x. \] Differentiate: \[ \frac{\partial}{\partial \theta} \ln f(x; \theta) = \frac{1}{\theta} + \ln x. \] Then, \[ I_1(\theta) = E\left[\left(\frac{1}{\theta} + \ln X\right)^2\right]. \]
Step 2: Compute expectation.
For \(f(x; \theta) = \theta x^{\theta - 1}\), \[ E(\ln X) = -\frac{1}{\theta}, \quad E((\ln X)^2) = \frac{2}{\theta^2}. \] Hence, \[ I_1(\theta) = \frac{1}{\theta^2}. \] For \(n\) samples, \(I_n(\theta) = \frac{n}{\theta^2}\).
Step 3: Cramer-Rao lower bound for \(\theta^3\).
If \(T\) is an unbiased estimator of \(\theta^3\), \[ \text{Var}(T) \ge \frac{(g'(\theta))^2}{I_n(\theta)} = \frac{(3\theta^2)^2}{n / \theta^2} = \frac{9\theta^6}{n}. \] Hence, (A) is TRUE.
Step 4: Analyze unbiasedness.
An unbiased estimator achieving equality in the CRLB requires a linear relationship between score and statistic, which is not possible for \(1/\theta\) in this case. Thus, no unbiased estimator of \(1/\theta\) attains the CRLB. Hence, (C) is TRUE. Final Answer: \[ \boxed{(A) \text{ and } (C)} \]