Discussion
Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample from \(N(\theta, 1)\) distribution. Which of the following is an MVUE of \(\theta^{2}\) ?
Answer: D
The MVUE of \(\theta^{2}\) is \(\overline{X}_{n}^{2} - \frac{1}{n}\).
\(\overline{X}_{n} \sim N(\theta, 1/n)\), so
\[ \begin{aligned} E(\overline{X}_{n}^{2}) &= \text{Var}(\overline{X}_{n}) + (E\overline{X}_{n})^{2} \\ &= \frac{1}{n} + \theta^{2} \end{aligned} \]
Subtracting \(1/n\) removes the bias. \(\overline{X}_{n}\) is complete and sufficient, and \(\overline{X}_{n}^{2} - 1/n\) is a function of it, so by the Lehmann-Scheffe theorem it is the minimum variance unbiased estimator.
\(\overline{X}_{n}^{2}\) is biased by \(1/n\). \(\overline{X}_{n}^{2} - 1\) over-corrects (it subtracts the variance of one observation, not of the mean), and \((\sum X_{i})^{2}/n = n\overline{X}_{n}^{2}\) has expectation \(1 + n\theta^{2}\).
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