Show the Statistic is Complete Consider a random sample $Y_1,\ldots,Y_n$ of the Uniform Distribution on the Interval $[-\phi,\phi]$
I'm wondering how I can show that the Statistic $$ T(\mathbf{Y}) = ( Y_{(1)} , Y_{(n)}) $$
is a Complete Statistic.
Thoughts so far : 
the pdf of $T$ can be represented as $$ f(x,y) = n(n-1) \left(\frac{y-x}{2 \phi} \right)^{n-2} \frac{1}{4 \phi^2}  \; \; \; \; - \phi < x < y < \phi $$
The expectation of any measurable function  of $T$ can be represented as $$E[g(T)] = \int_{- \phi}^{\phi} \int_{x}^{\phi} g(x,y) f(x,y) \,dy\,dx $$
Setting this equal to $0$ however does not really let me continue very far. I cant deduce completeness 
I can only guess that  I'm on the wrong track but I dont know how else to attempt this.
 A: Actually, $(Y_{(1)},Y_{(n)})$ is not a complete statistic for $U(-\phi,\phi)$.
Proof:
Method 1
If $(Y_{(1)},Y_{(n)})$ is complete statistic, since $(Y_{(1)},Y_{(n)})$ is sufficient it is also a minimal sufficient statistic by Bahadur's theorem.
We know that a minimal sufficient statistic for $U(-\phi,\phi)$ is $\max\{-Y_{(1)},Y_{(n)}\}$. Since minimal sufficient statistics are one-to-one corresponded, there should be a one-to-one map between $\max\{-Y_{(1)},Y_{(n)}\}$ and $(Y_{(1)},Y_{(n)})$ if $(Y_{(1)},Y_{(n)})$ is a minimal sufficient statistic. However, this is not possible since two different $(y_{(1)},y_{(n)})$ can have same $\max\{-y_{(1)},y_{(n)}\}$, e.g. when $(y_{(1)},y_{(n)})=(-1,0)$ or when $(y_{(1)},y_{(n)})=(0,1)$, $\max\{-y_{(1)},y_{(n)}\}$ are both $1$.
Method 2
We can write the p.d.f of $Y_{(n)}$ $$\delta_{Y_{(n)}}(y_n)=\frac{n}{(2\theta)^n}(y_n+\theta)^{n-1},y_n\in[-\phi,\phi]$$
and the p.d.f of $Y_{(1)}$ $$\delta_{Y_{(1)}}(y_1)=\frac{n}{(2\theta)^n}(\theta-y_1)^{n-1},y_1\in[-\phi,\phi].$$
Thus $$E_{\phi}[Y_{(n)}]=\frac{\phi(n-1)}{n+1}\text{ and }E_{\phi}[Y_{(1)}]=\frac{\phi(1-n)}{n+1}$$
This means if we let $g(x,z)=\frac{n+1}{n-1}z-\frac{n+1}{1-n}x$ we will get $$\mathbb{E}_{\phi}(g(Y_{(1)},Y_{(n)}))=0,\forall\phi\in\mathbb{R}\text{ but } g\not\equiv0,$$
which means $(Y_{(1)},Y_{(n)})$ is not complete.
