Not complete but minimal sufficient statistic Let $X =(X_1,\ldots, X_n), ~ X_i \mathrm{iid} \sim \mathcal N ( \theta, \theta^2), ~ \theta \in \Theta = \mathbb R \setminus \{0\}, ~ T(X)=(\sum_{i=1}^n X_i, \sum_{i=1}^n X_i^2)$.
I figured out that $T(X)$ is sufficient. To show that it's not complete I checked the function $g$ with $g(u,v) = 2u^2 - (n+1)v$ and noticed that $E_{\theta}[g(T(X))] = 0 ~ \forall \theta \in \Theta$ - but why is $P_{\theta}(g(T(X)) = 0) < 1$?
And how to show that $T(X)$ is minimal sufficient? I know that one can choose a subfamiliy $\mathcal P_0 \subset \mathcal P = \{f_{\theta} \mid \theta \in \Theta \}$ where $P$ is a family of densities with the same support and that's enough to show that $T$ is minimal sufficient for the subfamiliy. And if we have such a subfamily, $T^{*}(X) = \left(\frac{f_{\theta_1}(x)}{f_{\theta_0}(x)}, \ldots, \frac{f_{\theta_k}(x)}{f_{\theta_0}(x)}\right)$ is minimal sufficient (for $\Theta = \{\theta_0, \ldots, \theta_k\}$). But how to apply on the present case?

G(x) = (n+1)/2 ∑xi^2  – (∑xi )^2
E(G(x)) = 0 doesn't mean G(x)= 0
hence not complete
