Finding 2-dimensional sufficient statistic via Fisher-Neyman factorization when marginal pdf functions for x don't contain x

Question

Let $X_1,...,X_n$ be mutually independent with pdfs given by $f_i(X_i\mid\theta) = 1/(2i\theta) $ where $ -i(\theta - 1)<x_i<i(\theta +1) $ and $\theta>0.$ To find a two-dimensional sufficient statistic, I'm going to use Fisher-Neyman factorization.

The joint pdf of $X$ can be written as $$f(x|\theta)=1/((2\theta)^n\sum_i^n i) = (1- \bar x + \bar x)/((2\theta)^n\sum_i^n i)=(1- \bar x)/((2\theta)^n\sum_i^n i) + \bar x/((2\theta)^n\sum_i^n i) $$ over the domain of $X$. I need to factor $f(x|\theta)$ as $f(x|\theta)=h(x)g(T(x)|\theta)$, so letting $h(x)=1$, is $T=(T_1, T_2)=(\bar x,\sum_i^n i) $ a sufficient statistic? My concern is that $T_2$ doesn't involve $X$.

Answer 1

The support of the distribution depends on the parameter $\theta$. So use indicator functions for writing down the pdf correctly and hence get a sufficient statistic for $\theta$ using Factorization theorem.

First note that

$$-i\theta+i<x_i<i\theta+i\implies \theta>\max\left(\frac{x_i-i}{i},\frac{i-x_i}{i}\right)=\frac{|{x_i-i}|}{i}$$

So for $\theta>0$, joint density of $ X=(X_1,X_2,\ldots,X_n)$ is

\begin{align} f_{\theta}( x)&=\prod_{i=1}^n\frac{1}{2i\theta}\mathbf1_{-i(\theta-1)<x_i<i(\theta+1)} \\&=\frac{1}{n!(2\theta)^n}\mathbf1_{|{x_1-1}|,|x_2-2|/2,\ldots,|x_n-n|/n<\theta} \\&=\frac{1}{n!(2\theta)^n}\mathbf1_{\theta>\max\left(|x_1-1|,|x_2-2|/2,\ldots,|x_n-n|/n\right)} \end{align}

By the Factorization theorem, you get a sufficient statistic $$T(X)=\max_{1\le i\le n}\{|X_i-i|/i\}$$ for $\theta$. If you take any other statistic $T'$, then $(T,T')$ would also be sufficient for $\theta$ as $T'$ plays no further role in the amount of data reduction or condensation without losing information about $\theta$.