Show that the order statistics from an i.i.d sample from the family of all symmetric distributions on the line with known center of symmetry are not complete.
So without loss of generality, I can assume that the center of symmetry is along the origin by just reshifting the distribution. So let $f_\theta(x)$ be a distribution with symmetry along $x=0$.
Let $X_1, \cdots, X_n$ be an i.i.d sample from $f_\theta$ and $X_{(1)}, \cdots, X_{(n)}$ be the order statistics
I need to show that there exists some nonzero $\delta(X_{(1)}, \cdots, X_{(n)})$ so that $E_\theta\left[\delta(X_{(1)}, \cdots, X_{(n)}) \right] = 0$
But I'm unsure how to proceed. I would imagine that when integrating $\delta$ over a symmetric distribution, I should be able to exploit the symmetry to have $\delta \neq 0$ but the integral will be.
How about, for odd $n$, $\delta(X_{(1)},\ldots,X_{(n)})=\varphi(X_{((n+1)/2)})$, where $\varphi:\Bbb R\to\Bbb R$ is continuous, bounded and anti-symmetric. ($X_{((n+1)/2)}$, the sample median, has a distribution symmetric about $0$, because the distirbution of the order statistics is anti-invariant under reflection about $0$.) Something similar will work for even $n$.