2-dimensional minimal sufficient statistic for $U(-k\theta+k,k\theta+k)$

Question

Find a two dimensional minimal sufficient statistic for $\theta$ from $n$ independent random variables $X_k\sim > U(-k\theta+k,k\theta+k)$, $k\in\{1,\cdots,n\}$

Here is what I've attempted.

The pdf of X is $$\delta_\theta(x)=\prod_{i=1}^n\frac{1}{2k\theta}\mathbb{1}_{(1-\theta)k\leq x_k\leq(1+\theta)k}.$$ This means that $$\theta\ge\max\{1-\frac{x_k}{k},\frac{x_k}{k}-1\},\quad\forall k\in\{1,\cdots,n\}$$ or equivalently $|\frac{x_k}{k}-1|\leq\theta,\quad\forall k\in\{1,\cdots,n\}$. What I've got is $\underset{k\in\{1,\cdots,n\}}{\max}\left\{|\frac{x_k}{k}-1|\right\}$ as a minimal sufficient statistic. However, I am not sure if this is correct and the question was about there should be a 2-dimensional minimal statistic.

--Update It is always possible that the problem itself is wrong, i.e. there is no 2-dim minimal sufficient statistic. If this is the case, how to disprove it?

Answer 1

Your own attempted answer incorrectly treats $k$ as a fixed value, rather than an index. This gives you an incorrect likelihood function, which means that your subsequent work is also incorrect. We first observe the event equivalence:

$$-\theta k + k \leqslant x_k \leqslant \theta k + k \quad \quad \quad \iff \quad \quad \quad \Big| \frac{x_k}{k} - 1 \Big| \leqslant \theta,$$

which means that the correct likelihood function is:

$$\begin{align} L_\theta(\mathbf{x}_n) &= \prod_{k=1}^n \frac{1}{2k\theta} \cdot \mathbb{I}(\theta k - k \leqslant x_k \leqslant \theta k + k) \\[6pt] &= \prod_{k=1}^n \frac{1}{2k\theta} \cdot \mathbb{I} \Bigg( \theta \geqslant \Big| \frac{x_k}{k} - 1 \Big| \Bigg) \\[6pt] &= \frac{1}{(2\theta)^n n!} \cdot \mathbb{I} \Bigg( \theta \geqslant \max_k \Big| \frac{x_k}{k} - 1 \Big| \Bigg). \\[6pt] \end{align}$$

Consequently, a minimal sufficient statistic for the parameter $\theta$ is:

$$\max_k \Big| \frac{x_k}{k} - 1 \Big| .$$