Product of Indicator Functions
Theorem
For $x_{1} , \cdots , x_{n} \in \mathbb{R}$ and constant $\theta \in \mathbb{R}$, the product of $I_{\cdot} \left( x_{i} \right)$ is as follows: $$ \prod_{i=1}^{n} I_{[\theta,\infty)} \left( x_{i} \right) = I_{[\theta,\infty)} \left( \min_{i \in [n]} x_{i} \right) $$
- $I_{A}$ is the indicator function for the set $A$. $$ I_{A} (x) = \begin{cases} 1 & , x \in A \\ 0 & , x \notin A \end{cases} $$
Proof
Regardless of how many $x_{i}$ are in $[\theta , \infty)$, if the smallest $\min x_{i}$ is less than $\theta$, it eventually leads to $0$, and the rest is the product of $1$, so it is not necessary to consider all $x_{i}$.
■
Explanation
Reverse Direction
It is necessary for the proof of the theorem related to sufficient statistics. Although it’s obvious, one can consider the following theorem in the opposite direction: $$ \prod_{i=1}^{n} I_{(-\infty, \theta]} \left( x_{i} \right) = I_{(-\infty, \theta]} \left( \max_{i \in [n]} x_{i} \right) $$
In the case where $x$ is fixed
The introduced theorem considered $x_{1} , \cdots , x_{n} \in \mathbb{R}$ as a variable and the set was fixed. On the other hand, when $x$ is fixed and $A_{1} , \cdots , A_{n}$ is variable, one can consider the product of indicator functions as follows. $$ \prod_{i=1}^{n} I_{A_{i}} (x) = I_{\bigcap_{i=1}^{n} A_{i}} (x) $$