0
$\begingroup$

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.

$\endgroup$
0
2
$\begingroup$

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.


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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.