# Sufficient statistic of $f(\theta;x)=\theta x^{-2}I_{[\theta,\infty)}(x)$

Let $X_1,\dots,X_n$ be a random sample from the density

$f(\theta;x)=\theta x^{-2}I_{[\theta,\infty)}(x)$

where $\theta>0$. Is $X_{(1)}=\min\left[X_i\right]$ a sufficient statistic?

My attempt

I made the joint density function like so:

\begin{align} L\left(\theta; x_1,\dots, x_n \right)&=\prod_{i=1}^n\theta x_i^{-2}I_{[\theta,\infty)}(x_i) \\ &= \theta^n\prod_{i=1}^nx_i^{-2}\prod_{i=1}^nI_{[\theta,\infty)}(x_i) .\end{align}

Now I want to rewrite $\prod_{i=1}^nI_{[\theta,\infty)}(x_i)$ in terms of $X_{(1)}$, but I'm not sure how to do. Any suggestions?

Observe that $$\prod_{i=1}^n \mathbb 1(x_i \ge \theta) = \mathbb 1\left( \bigcap_{i=1}^n x_i \ge \theta \right) = \mathbb 1 \left(\min_i x_i \ge \theta\right) = \mathbb 1 (x_{(1)} \ge \theta).$$