Find a one-dimensional sufficient statistic for $\theta$ given that $f(x;\theta)=\frac{1}{\theta^2}x e^{-\frac{x}{\theta}} I_{(0,\infty)}(x)$ Assume that $(X_1, X_2, X_3, \dots, X_n)$ is a random sample of the distribution having the following probability distribution function (PDF):
$$f(x;\theta)=\frac{1}{\theta^2}x e^{-\frac{x}{\theta}} I_{(0,\infty)}(x), \quad \theta >0$$
where $I_{(0,\infty)}(x)$ is an indicator function on the set $(0,\infty)$ (meaning that if $x$ belongs to that set, the output of the function is $1$, otherwise it's $0$).
Question: Find a one-dimensional sufficient statistic for $\theta$.
I'm stuck at many things:
(1) What is a "one-dimensional" sufficient statistic? I know that a sufficient statistic somehow summarizes the data such that if we only see that statistic, we will do the same as if the real data was shown to us.
(2) Assuming that we know what a one-dimensional sufficient statistic for $\theta$ is, how should I proceed further? I mean, is there a systematic way to find a sufficient statistic? or we just guess it and try to prove that it is actually a sufficient statistic?
Note: I also saw this post which seems relevant. However, I am not sure how to use the factorization theorem (if it can be useful in my case!). What are the factors here? And how should I get rid of the indicator function?
 A: 
What is a "one-dimensional" sufficient statistic?

One-dimensional sufficient statistic is a scalar function $T(X)$ of the sample $X= (X_1,...,X_n)$ such that the Fisher information in $T(X)$ equals the Fisher information in $X$.

Assuming that we know what a one-dimensional sufficient statistic for θ is, how should I proceed further? I mean, is there a systematic way to find a sufficient statistic? or we just guess it and try to prove that it is actually a sufficient statistic?

Yes, there is a systematic way to find a sufficient statistic, that is the factorization criterion. Namely, if you can factorize the likelihood function into two multiplicative terms such that one depends on the unknown parameter $\theta$ only through the $T(X)$ and the other depends only on known terms, then $T(X)$ is the sufficient statistic for the unknown parameter $\theta$. For further details please refer to
https://en.wikipedia.org/wiki/Sufficient_statistic#Fisher%E2%80%93Neyman_factorization_theorem
For your density function - by using the factorization criterion we get
\begin{align}
\prod_{i=1}^n f(x_i ; \theta) = \frac{1}{\theta^{2n}}e^{-\frac{1}{\theta}\sum_{i=1}^n x_i} \times \prod x_i I_{(0, \infty)}(x_i),
\end{align}
thus the one-dimensional sufficient statistics is
$$
T(X) = \sum_{i=1}^n x_i
$$
A: A sufficient statistic $T(X)$ for a specified family of probability distributions of random data $X$ is a function $T$ with the property that the conditional distribution of $X$ given $T(X)$ is the same for all distributions in the specified family.
Thus you seek a function $T(X_1,\ldots,X_n)$ for which the conditional distribution of $(X_1,\ldots,X_n)$ given $T(X_1,\ldots,X_n)$ remains the same as $\theta$ changes.
"One-dimensional" means the value of $T$ is a scalar, i.e. a real number, rather than an ordered pair or something else besides just a single number.
The joint density is
\begin{align}
f(x_1,\ldots,x_n) & = \prod_{i=1}^n \frac{1}{\theta^2}x_i e^{-x_i/\theta} \quad \text{for } x_1,\ldots,x_n>0 \\[8pt]
& = \left( \prod_{i=1}^n x_i \right) e^{-(x_1+\cdots+x_n)/\theta} \cdot \theta^{-2n}
\end{align}
This has one factor (remember: "factors" are things that are multiplied) that does not depend on $\theta$ and and one that depends on $x_1,\ldots,x_n$ only through $x_1+\cdots+x_n.$
Therefore, by Fisher's factorization theorem, $T(X_1,\ldots,X_n) = X_1+\cdots+X_n$ is a sufficient statistic for this family of distributions.
