Doubt Regarding the Sufficient Statistic Problem

Question

Let $X_1, X_2, ..., X_n$ be random variables from the following densities: $$ f(x_i|\theta) = \frac{1}{2i\theta} \text{ for } -i(\theta-1) < x_i < i(\theta+1) $$ and otherwise the density is zero.

This is a problem from Casella and Berger statistical inference second edition that requires to produce a two dimensional sufficient statistic.

The answer says that $$\left(\frac{\min(x_i)}{i}, \frac{\max(x_i)}{i}\right)$$ will be a two dimensional sufficient statistic. My question is that why not $$\left\{-\min([x_i-i]/i), \max([x_i-i]/i)\right\}$$ will be a sufficient statistic?

I think it will and we can use the one one function invariance property of sufficient statistic for that? I am sure if I am thinking in the right direction.

Answer 1

Note: there was a typo in the question (but not in the original exercise) that I did not want to correct as it would have changed the meaning of the question: $$\left(\frac{i-\min(x_i)}{i}, \frac{\max(x_i)-i}{i}\right)$$ was replaced with $$\left\{-\min([x_i-i]/i), \max([x_i-i]/i)\right\}$$ as the extrema applies to all terms involving $i$ at once.

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The likelihood writes as $$L(\theta|x_1,\ldots,x_N) = \prod_{i=1}^n \frac{1}{i\theta} \mathbb I_{(-i\theta+i,i\theta+i)}(x_i)$$ and $$\prod_{i=1}^n \mathbb I_{(-i\theta+i,i\theta+i)}(x_i) = \prod_{i=1}^n \mathbb I_{((i-x_i)/i,\infty)}(\theta) \times \prod_{i=1}^n \mathbb I_{((x_i-i)/i,\infty)}(\theta)$$ means that the likelihood is zero unless $$\theta\ge\max_i(1-x_i/i)=1-\min_i(x_i/i)\quad\text{and}\quad\theta>\max_i(x_i/i-1)$$ i.e. $$\theta\ge\max\{1-\min_i(x_i/i),\max_i(x_i/i)-1\}$$ which is a function of $$\{\min_i(x_i/i),\max_i(x_i/i)\}$$ and a function of $$\{\min_i([x_i-i]/i),\max_i([x_i-i]/i)\}$$ equivalently. (The answer to the question boils down to $i/i=1$ being superfluous.)