Prove $(\sum_{i=1}^{n}X_{i},\sum_{i=1}^{n}X_{i}^{2})$ is not a complete statistic for $N(\mu,\mu^2)$ distribution

Question

Let $X_{1},\ldots,X_{n}\stackrel{\text{ i.i.d }}{\sim}N(\mu,\mu^{2})$.

$T=\left(\sum_{i=1}^{n}X_{i},\sum_{i=1}^{n}X_{i}^{2}\right)$ is a sufficient statistic for $\mu$. Also $T$ is minimal sufficient.

However, the question is: Prove that $T$ is not complete, hint: use $f(x,y)=x-2y^{2}/(n+1)$.

To prove that $T$ is not complete, I must find an example of a Borel function, $f$, for which $E_{\theta}(f(T))=0$ does not imply $f(T)=0$. However, I do not know how to interpret the 'hint', since I only can conclude that $f(T)=0$, whenever using $x=\sum_{i=1}^{n}X_{i}$ and $y=\sum_{i=1}^{n}X_{i}^{2}$ and calculating $E_{\theta}(f(T))=0$.

Answer 1

Consider the function $g$ of the sufficient statistic $T$ defined as $$g(T(X_1,\ldots,X_n))=2\left(\sum_{i=1}^n X_i\right)^2-(n+1)\sum_{i=1}^nX_i^2$$

For all possible values of $\mu$, verify that we have

$$E_{\mu}g(T(X_1,\ldots,X_n))=2n(1+n)\mu^2-(n+1)2n\mu^2=0$$

However, $g(T)$ is not identically zero with probability one.

So the minimal sufficient statistic $T$ is not complete.