Sufficient statistic for Double Exponential Let $X_1,X_2,...X_n$ be a random sample from  $f(x,\theta)=\frac{1}{2 \theta}e^{\frac{-|x|}{\theta}}$.We know by Factorisation theorem that $\frac{\sum |X_i|}{n}$ is sufficient for $\theta$.
But can we show that $\frac{\sum X_i}{n}$ is not sufficient for $\theta$?
It is tough to show it by definition as the distribution of $\sum X_i$ cannot be found explicitly.
Is there any other way?

let $S$ is minimal sufficient. $T$ is not sufficient if exist $x,y\in support$ 
$T(x)=T(y)$ but $S(x)\neq S(y)$ 
let n=2, $T(x)=x_1+x$ and $S(x)=|x_1|+|x_2|$ (S is minimal sufficient) 
$x=(2,-1)$ and $y=(3,-2)$
$T(x)=1=T(y)$ $S(x)=3\neq 5=T(y)$
so $T$ is not sufficient
in general for arbitrary $n$ choose  $x=(2,-1,0,\cdots ,0)$ $y=(3,-2,0,\cdots ,0)$
this method is based of this point that minimal sufficient is a function of any sufficient Statistic, and in above we shown that $S$ is not a function of $T$.note $V$ is a function of $U$ if
$\forall x,y \quad U(x)=U(y) \Longrightarrow V(x)=V(y)$ so 
$V$ is not a function $U$ if
$\exists x,y \quad V(x)=V(y) \quad but \quad U(x)\neq U(y)  $
