Intuition for why $X_{(1)}$ is a minimal sufficient statistic

Question

Let $$X_i \sim f(x;\theta) = \begin{cases}1.5e^{1.5(\theta-x)}\quad \text{if} \ x\geq 0, \\ 0\quad\quad\quad \qquad \text{else.}\end{cases}$$ be iid for $i=1,2,\ldots,n$ and $\theta$ is an unknown parameter.

I can easily show that $X_{(1)}$ will be minimal sufficient just by using Lehmann-Scheffe's Theorem, but the question says to "justify" which I interpret that there is probably a simple/intuitive explanation.

Answer 1

This is a shifted exponential distribution that takes $\theta$ as its minimum possible value; $\theta$ is also the mode. So the minimum observed value $X_{(1)}$ is obviously the best clue as to the value of $\theta.$

Graph of the density function for $\theta = 2.$

Here is a histogram of a sample of size $n = 100$ from this distribution. Tick marks show locations of individual observations.

Summary statistics for this sample are as shown below. The smallest observation $X_{(1)} = 2.002$ comes very close to the minimum possible value $\theta = 2.$

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.002   2.179   2.398   2.648   2.759   8.818 

For more information see this Q&A.