Question

If \(t_n\) is an efficient estimator of the population mean \(\mu\), then :

• A. Var(\(t_n\)) tends to 0 as \(n \to \infty\)
• B. Var(\(t_n\)) becomes equal to \(\sigma^2\) for large sample size n
• C. \(t_n\) converges to \(\mu\) as \(n \to \infty\)
• D. None of these

Hint:

An efficient estimator's variance equals the Cramer-Rao bound \(1/(nI(\theta))\), which shrinks as n grows.

Answer 1

The Correct Option is A

Step 1: An efficient estimator is one whose variance attains the Cramer-Rao lower bound, that is \(\text{Var}(t_n) = \dfrac{1}{n I(\theta)}\), where \(I(\theta)\) is the Fisher information in a single observation.

Step 2: As the sample size \(n\) grows, the denominator \(n I(\theta)\) grows without bound, so \(\text{Var}(t_n) = \dfrac{1}{n I(\theta)} \to 0\) as \(n \to \infty\).

Step 3: For instance, the sample mean is the efficient estimator of \(\mu\) for a normal population, with \(\text{Var}(\bar{x}) = \sigma^2 / n\), which clearly tends to 0, not to \(\sigma^2\), as n grows.

Step 4: So the variance of an efficient estimator of the mean shrinks to zero with increasing sample size, matching option (A) exactly.

Final Answer: (A) Var(\(t_n\)) tends to 0 as \(n \to \infty\).