I need to understand an elementary part of sufficient statistics.
$X_1,X_2,\ldots,X_n \space$ are a random sample.
Let
\begin{align}
& (i) & & X_1,X_2,\ldots,X_n \sim U(0, \theta), \ \theta>0 \\
& (ii) & & X_1,X_2,\ldots,X_n \sim U\left(\theta-\frac{1}{2}, \theta+\frac{1}{2}\right), \ \theta>0
\end{align}
In both the cases the problem is to find the sufficient statistic for $\theta$.
Applying the factorization theorem, from $(i)$, we have
\begin{align}
f(\mathbf{x},\theta) & = \begin {cases} \frac{1}{\theta^n} & \text{if } 0<x_i<\theta, \space \space i=1,2,\ldots,n\\ 0 & \text{otherwise}\end{cases}\bigg\}\\
& = \underbrace{\frac{1}{\theta^n}\space \text{I}_{(\text{x}_{(n)})}^{(0,\theta)}}_{\text{g}\textbf(x_{(n)},\theta)}\space \underbrace{\prod_{i=1}^{n-1} \text{I}_{(\text{x}_{(i)})}^{(0,x_{(n)}}}_\text{h(x)}
\end{align}
From $(ii)$, we have
\begin{align}
f(\mathbf{x},\theta) & =\begin {cases} 1 & \text{if } \theta – \frac{1}{2}<x_{(1)}<x_{(2)}<\cdots<x_{(n)}<\theta+\frac{1}{2}, \space \space i=1,2,\ldots,n\\ 0 & \text{otherwise}\end{cases}\bigg\}\\
& =\underbrace{\text{I}_{(\text{x}_{(1)})}^{(\theta – \frac{1}{2} ,\theta+\frac{1}{2})}\space \text{I}_{(\text{x}_{(n)})}^{(\theta – \frac{1}{2} ,\theta+\frac{1}{2})}}_{\text{g} \textbf(x_{(1)},x_{(n)},\theta)} \space \underbrace{\prod_{i=2}^{n-1} \text{I}_{(\text{x}_{(i)})}^{(x_{(1)},x_{(n)})}}_\text{h(x)}
\end{align}
And from $(i)$ $x_{(n)}$ is the sufficient statistic for $\theta$ and from $(ii)$ $(x_{(1)},x_{(n)})$ is the sufficient statistic for $\theta$.
What I don't understand –
In problem $(ii)$ $g(\mathbf{x},\theta) = \text{I}_{(\text{x}_{(1)})}^{(\theta – \frac{1}{2} ,\theta+\frac{1}{2})}\space \text{I}_{(\text{x}_{(n)})}^{(\theta – \frac{1}{2} ,\theta+\frac{1}{2})}. \prod_{i=2}^{n-1}\text{I}_{(\text{x}_{(i)})}^{(x_{(1)},x_{(n)})}$ then similarly in problem $(i)$ the $\mathcal g$(x,$\theta)$ should be equal to $\text{I}_{(\text{x}_{(1)})}^{(0,\theta)}\space \text{I}_{(\text{x}_{(n)})}^{(0,\theta)}. \prod_{i=2}^{n-1}\text{I}_{(\text{x}_{(i)})}^{(x_{(1)},x_{(n)})}$
Why is not $x_{(1)}$ is an sufficient statistic in the problem $(i)$
When $X_1,X_2,\ldots,X_n \space \sim U(\frac{-\theta}{2}, \frac{\theta}{2})$ the sufficient statistic is same as the problem $(ii)$, whereas the pdf is same as the problem $(i)$
I need to understand the reason that having same pdf why sufficient statistics change or having different pdf how come the sufficient statistic become similar when order statistics is involved. Any help, explanation is valuable and highly appreciated.
Best Answer
In $(i),$ the pair $(x_{(1)}, x_{(n)})$ is in fact sufficient, but it is not a minimal sufficient statistic, since a coarser statistic, $x_{(n)}$ is also sufficient. ("$A$ is coarser than $B$" would mean given $B$ you can compute $A$ but given $A$ you cannot compute $B$ without further information.)