Discussion
Suppose \(X_{1}, \ldots, X_{n}\) are independent random variables with probability density function
\[ f(x_{i}, \theta) = \begin{cases} \dfrac{1}{2i\theta}, & \text{if } -i\theta \lt x_{i} \lt i\theta \\ 0, & \text{otherwise.} \end{cases} \]
Then the minimal sufficient statistic for \(\theta\) is :
Answer: B
The minimal sufficient statistic is \(\max\{|X_{1}|, |X_{2}|/2, \ldots, |X_{n}|/n\}\).
The joint density is
\[ \begin{aligned} &\prod_{i} \frac{1}{2i\theta}1_{\{|x_{i}| \lt i\theta\}} \\ &= \frac{1}{(2\theta)^{n} n!}1_{\{\max_{i} |x_{i}|/i \lt \theta\}} \end{aligned} \]
which depends on the data only through \(T = \max_{i}|X_{i}|/i\). By factorisation T is sufficient, and since the likelihood as a function of \(\theta\) is determined exactly by T, it is minimal.
The order statistics in (A) are sufficient but not minimal (and ignore that the \(X_{i}\) have different scales). Options C and D drop the absolute values (as printed) or take the minimum, which loses the information that matters.
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