Exercise :
Let $X_1, \dots, X_n$ be a random sample from a distribution with probability density function $f(x) = \theta x^{-2}, \; \; 0 < \theta \leq x < \infty$, where $\theta$ unknown parameter.
(i) Check if the given distribution belongs to the Exponential Family of Distributions.
(ii) Find a sufficient statistic function $T$ of $\theta$.
Attempt :
(i) It is :
It does not belong to the Exponential Family of Distributions, since the support of $X$ depends on $\theta$.
(ii) To try and find a sufficient statistic function, we must write $p(x|\theta)$ in the form of $G(t,\theta)H(x)$.
Thus :
$$p(x|\theta) = \prod_{i=1}^n \theta x_i^{-2} \mathbb{I}_{[\theta, + \infty]}(x_i)$$
How should I continue to find a sufficient statistic function from here on?
Using the factorization criteria you have $$ \prod_{i=1}^n \theta x_i^{-2}I[x_i\ge \theta]=\theta^nI[x_{(1)}\ge\theta]\times \prod_{i=1}^n\frac{1}{x_i^2}I[x_i\ge x_{(1)}]=g(\theta ; T(X))H(X), $$ i.e., the MSS is $T(X)=X_{(1)}=\min\{X_1,...,X_n\}$. It is minimal as it one dimensional like $\theta$. The last step follows from the fact that $\prod_{i=1}^nI[x_i \ge \theta] = I[x_{(1)} \ge \theta]\prod_{i=2}^nI[x_i \ge x_{(1)}]$. And note that as the support of $X$ depends on $\theta$, thus it cannot belong to the exponential family.
Another approach is to verify that $X_{(1)}$ is the MLE, hence it must be a function of the sufficient statistics, and as $X_{(1)}$ is the identity function and it is one dimensional then it is the minimal sufficient statistic.