Question

If $X_{1}, X_{2}, \dots, X_{n}$ denote a random sample of size $n$ from normal population $N(0, \theta^{2})$ then MVUE of $\theta^{2}$ is

• A. $\overline{x}^{2}$
• B. $\frac{1}{n} \sum_{i=1}^{n} X_{i}^{2}$
• C. $\sum_{i=1}^{n} \frac{(X_{i}-\overline{X})^{2}}{n-1}$
• D. $\sum_{i=1}^{n} \frac{(X_{i}-\overline{X})^{2}}{n}$

Hint:

When the mean $\mu$ is known (e.g., $\mu=0$), we do not lose a degree of freedom. In such cases, the divisor for the unbiased estimator of variance is $n$, whereas it is $n-1$ when the mean is unknown and estimated by $\overline{X}$.

Answer 1

The Correct Option is B