Question

For \( n \geq 2 \), let \( \epsilon_1, \epsilon_2, \ldots, \epsilon_n \) be i.i.d. random variables having the \( N(0,1) \) distribution. Consider \( n \) independent random variables \( Y_1, Y_2, \ldots, Y_n \) defined by \( Y_i = \beta + \epsilon_i \), \( i = 1,2, \ldots, n \), where \( \beta \in \mathbb{R} \). Define
\[ \bar{Y} = \frac{1}{n} \sum_{i=1}^{n} Y_i \]
\[ T_1 = \frac{2\bar{Y}}{n+1} \]
\[ T_2 = \frac{1}{n} \sum_{i=1}^{n} \frac{Y_i}{i} \]
Then which of the following statements is NOT correct?

• A. \( T_1 \) is an unbiased estimator of \(\beta\)
• B. \( T_2 \) is an unbiased estimator of \(\beta\)
• C. \(\mathrm{Var}(T_1) < \mathrm{Var}(T_2)\)
• D. \(\mathrm{Var}(T_1) = \mathrm{Var}(T_2)\)

Answer 1

The Correct Option is D

1. Unbiasedness: - \( T_1 = \frac{2\bar{Y}}{n+1} \), where \( \bar{Y} \) is an unbiased estimator of \( \beta \). Hence, \( T_1 \) is also unbiased. - \( T_2 = \bar{Y} \), which is the sample mean, is an unbiased estimator of \( \beta \). 2. Variance Comparison: - The variance of \( T_1 \) is given by: \[ \text{Var}(T_1) = \frac{4 \cdot \text{Var}(\bar{Y})}{(n+1)^2} = \frac{4 \cdot \frac{1}{n}}{(n+1)^2}. \] - The variance of \( T_2 \) is: \[ \text{Var}(T_2) = \text{Var}(\bar{Y}) = \frac{1}{n}. \] - Clearly, \( \text{Var}(T_1) < \text{Var}(T_2) \). 3. Equality of Variance: - Since \( \text{Var}(T_1) \neq \text{Var}(T_2) \), statement (D) is incorrect