Let $X_1,...,X_n$ be iid and uniformly distributed on the interval $(-\theta, \theta)$ where $\theta>0$. I am asked to derive the set of minimal sufficient statistics. Here's my work:
For each $X_i$, the pdf is given by $f(x)=\frac{1}{2\theta}I(-\theta<x_i<\theta).$ Hence, the joint pdf for $X$ can be given by $g(x)=(\frac{1}{2\theta})^nI(-\theta<x_i<\theta)$ for all $i=1, 2,...,n$.
Now let $h(x,y)=\frac{g(x)}{g(y)}$ for a particular combination of $X$ and $Y$ belonging to the sample space. Then $h(x,y)$ can be rewritten as follows:$$h(x,y)=\frac{(\frac{1}{2\theta})^nI(-\theta<x_i<\theta)}{(\frac{1}{2\theta})^nI(-\theta<y_i<\theta)}=\frac{I(-\theta<x_i<\theta)}{I(-\theta<y_i<\theta)}=\frac{I(max(|x_i|)<\theta)}{I(max(|y_i|)<\theta)} $$ I know that a statistic $T$ is minimally sufficient for $\theta$ if the following holds: $h(x,y)$ is independent of $\theta$ for all possible values of $\theta$ if and only if $T(x)=T(y)$.
My question is as follows: if $max(|x_i|)=max(|y_i|)$, would my final expression above for $h(x,y)$ be free of $\theta$ for all possible values of $\theta$? I want to say that $h(x,y)$ would equal 1 for all $\theta$, but couldn't this expression also be undefined? Thanks!