Finding complete sufficient statistic Let $X_1, \dots, X_n$ be iid. $\text{Uniform}[-\theta,\theta]$. I need to find the complete sufficient statistic for $\theta$ or prove there does not exist such.
I know that $T = (X_{(1)}, X_{(n)} )$ is a sufficient statistic for $\theta$ but it is not a complete sufficient statistic.
I want to prove it. So first I tried to use the Basu's theorem . But in this case $R = X_{(n)} - X_{(1)}, $ is not an ancillary statistic.
So I tried prove using the definition of the complete sufficient statistic.
Here I have attached my work so far:

But by doing like this , seems like that I am going to prove that $T$ is a complete sufficient statistic.
So can someone help to figure it out what I did incorrectly ?
 A: Recall:

Definition: A statistic $T$ is complete for $\theta$ if $$E(g(T)) = 0, \ \text{ for all $\theta$}  \quad \Rightarrow \quad P(g(T) = 0) = 1, \ \text{ for all $\theta$}$$

The part about $P(g(T) = 0) = 1$ basically says that the function $g$ is trivially $0$ everywhere (except possibly on a set of measure 0).
So... If you want to prove that $T$ is NOT complete, you can try to find a non-trivial function $g(T)$ for which $E(g(T)) = 0$ for all values of $\theta$.
Hint: Can you find $E(X_{(1)})$ and $E(X_{(n)})$? Start with that, and then try looking at linear combinations of the sufficient order statistics.
A: Method 1
$(X_{(1)},X_{(n)})$ is not complete because we can find $g\neq0$ but $\mathbb{E}\left[g(X_{(1)},X_{(n)})\right]=0,\forall\theta$. $g$ is $(t_1,t_2)\rightarrow\frac{n+1}{n-1}t_2-\frac{n+1}{1-n}t_1$.
This is because  $\mathbb{E}(X_{(n)})=\frac{n-1}{n+1}\theta$ and $\mathbb{E}(X_{(1)})=\frac{1-n}{n+1}\theta$. Thus $\mathbb{E}\left[g(X_{(1)},X_{(n)})\right]=\mathbb{E}\left[\frac{n+1}{n-1}X_{(n)}-\frac{n+1}{1-n}X_{(1)}\right] = \frac{n+1}{n-1}\mathbb{E}(X_{(n)})-\frac{n+1}{1-n}\mathbb{E}(X_{(1)}) = \theta-\theta=0,\forall \theta$.
Method 2
If the sufficient statistic $(X_{(1)},X_{(n)})$ is complete, then it is a minimal sufficient statistic. However, (X_{(1)},X_{(n)}) is not a minimal sufficient statistic. A minimal sufficient statistic is $\max\{-X_{(1)},X_{(n)}\}$. It is possible that $(x_{(1)},x_{(n)})\neq(y_{(1)},y_{(n)})$ but $\max\{-x_{(1)},x_{(n)}\}=\max\{-y_{(1)},y_{(n)}\}$.
