Discussion
Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. random variables such that \(X_{1}\) has normal distribution with mean \(\theta\) and variance \(\theta \gt 0\). Then, which of the following is a minimal sufficient statistic for \(\theta\) ?
Answer: D
The minimal sufficient statistic is \(\sum_{i=1}^{n} X_{i}^{2}\).
With mean and variance both \(\theta\), the joint density is
\[ \begin{aligned} &(2\pi\theta)^{-n/2}\exp\left(-\frac{\sum (x_{i} - \theta)^{2}}{2\theta}\right) \\ &= (2\pi\theta)^{-n/2}\exp\left(-\frac{\sum x_{i}^{2}}{2\theta} + \sum x_{i} - \frac{n\theta}{2}\right) \end{aligned} \]
The term \(e^{\sum x_{i}}\) does not involve \(\theta\), so it factors out, and the data enter the \(\theta\)-dependent part only through \(\sum x_{i}^{2}\). It is a one-parameter exponential family in \(\sum x_{i}^{2}\), which is minimal sufficient.
\(\overline{X}\) and \(\sum (X_{i} - \overline{X})^{2}\) alone are not sufficient. The pair in option C is sufficient (it determines \(\sum X_{i}^{2}\)) but not minimal, since it carries more than is needed.
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