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?

  • $\begingroup$ The value $k$ is an index, not a fixed quantity, so it should not appear in your answer. $\endgroup$
    – Ben
    Jan 5 at 0:45
  • $\begingroup$ Updated. Thanks. @Ben $\endgroup$
    – Tan
    Jan 5 at 0:49
  • $\begingroup$ Similar: stats.stackexchange.com/q/510173/119261 $\endgroup$ Feb 19 at 19:06

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:

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

  • 2
    $\begingroup$ But what is 'two dimensional' about this sufficient statistic as asked in the question? $\endgroup$ Jan 5 at 10:36
  • $\begingroup$ It is not two-dimensional. Is it possible you have misstated the bound in the problem? (I ask because if you change the lower bound to $-k + k \theta$ then you get a two dimensional minimum sufficient statistic instead.) $\endgroup$
    – Ben
    Jan 5 at 21:16
  • $\begingroup$ No, I double-checked.@Ben It is indeed $-k\theta+k$. I would be interesting then to show there is no two-dimensional minimal sufficient statistic. I don't think there is any theorem about the smallest dimension of minimal sufficient statistics. $\endgroup$
    – Tan
    Jan 7 at 5:47
  • $\begingroup$ On the contrary, if you show that you have a scalar statistic that is sufficient, then that is proof that there is no (non-degenerate) two-dimensional statistic that is also minimal sufficient. It is a fairly simple application of the definition of minimal sufficiency. $\endgroup$
    – Ben
    Jan 7 at 6:03

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