A curved normal family can make a sufficient statistic incomplete

The map $T$ is polynomial and hence a statistic by [F2]. Let ${\lambda }_B$ be one-dimensional Lebesgue measure restricted to $\mathcal{B}(\mathbb{R})$; the Borel half-open intervals $(-k,k]$ give it a sigma-finite cover of finite measure by [F5]. By [F4] and the affine change of variables [F10] for $z\mapsto \theta +\theta z$, each $X_i$ has Borel density $q_{\theta }(x_i)=(\sqrt{2\pi }\theta {)}^{-1}{e}^{-(x_i-\theta {)}^2\mathrm{/}(2{\theta }^2)}$ relative to ${\lambda }_B$. Independence [F5] gives the joint law $P_{\theta }^{X_1}\times P_{\theta }^{X_2}$. The measure with density $q_{\theta }(x_1)q_{\theta }(x_2)$ relative to ${\lambda }_B\times {\lambda }_B$ exists by [F9]; Tonelli [F6] shows that it has total mass one and agrees with this product law on every measurable rectangle. By [F11], these rectangles generate the Borel sigma-algebra of ${\mathbb{R}}^2$, so the product-measure uniqueness clause [F9] identifies this density measure with the joint law. Expanding its density gives $p_{\theta }(x_1,x_2)=(2\pi {)}^{-1}{\theta }^{-2}\exp (-S_2\mathrm{/}(2{\theta }^2)+S_1\mathrm{/}\theta -1)$. Put $u=1\mathrm{/}\theta$, $h(x_1,x_2)=(2\pi {)}^{-1}$, $\mathrm{\Omega }=\{(u,-u^2\mathrm{/}2):u>0\}$, and $A(u,-u^2\mathrm{/}2)=1-2\log u$. Then $p_{\theta }=h\exp \{u{S}_1-u^2{S}_2\mathrm{/}2-A(u,-u^2\mathrm{/}2)\}$, with normalizer $Z(u,-u^2\mathrm{/}2)=e^A=e\mathrm{/}{u}^2$; the normalization follows from the product normal density just derived. Thus the canonical statistic is $(S_1,S_2)$, and the natural parameter set lies on ${\eta }_2=-{\eta }_1^2\mathrm{/}2$, so it contains no open subset of ${\mathbb{R}}^2$. Both densities are strictly positive, so $P_{\theta }\ll P_{{\theta }_0}$ for any fixed ${\theta }_0>0$. Their ratio is a Borel function of $T$; since $P_{{\theta }_0}$ is a model member, [F7] proves sufficiency.

From $X_i=\theta +\theta Z_i$ and [F4], $E_{\theta }{X}_i=\theta$ and $E_{\theta }{X}_i^2=2{\theta }^2$. The bound $\mathrm{\mid }z\mathrm{\mid }\le 1+z^2$ gives $E\mathrm{\mid }{Z}_i\mathrm{\mid }\le 1+E{Z}_i^2=2$, hence $E_{\theta }\mathrm{\mid }{X}_i\mathrm{\mid }<\mathrm{\infty }$. Let $q_{\theta }$ be the one-dimensional normal density from step 1.1 and write $x^{+}=\max (x,0)$, $x^{-}=\max (-x,0)$. Independence [F5] identifies the joint law with the product of the two marginal laws, each having density $q_{\theta }$. Tonelli [F6] therefore gives, for every $a,b\in \{+,-\}$, $E_{\theta }[X_1^a{X}_2^b]=({\int }_{\mathbb{R}}{x}^a{q}_{\theta }(x)dx)({\int }_{\mathbb{R}}{x}^b{q}_{\theta }(x)dx)$, where all four factors are finite by $E_{\theta }\mathrm{\mid }{X}_i\mathrm{\mid }<\mathrm{\infty }$. Subtracting the four finite identities yields $E_{\theta }[X_1{X}_2]=(E_{\theta }{X}_1)(E_{\theta }{X}_2)$; linearity [F8] now gives $E_{\theta }{S}_1^2=E{X}_1^2+E{X}_2^2+2E{X}_{1}E{X}_2=6{\theta }^2$, while $E_{\theta }{S}_2=4{\theta }^2$. Hence $E_{\theta }g(T)=6{\theta }^2-\frac{3}{2}(4{\theta }^2)=0$ for every $\theta >0$. Also $\mathrm{\mid }g(T)\mathrm{\mid }\le S_1^2+\frac{3}{2}{S}_2$, whose expectation is finite, so $g(T)$ is integrable.

At $(x_1,x_2)=(1,0)$, $g(T)=-1\mathrm{/}2$. By continuity it is nonzero on a neighborhood. The density in step 1.1 is continuous and strictly positive at this point, so there is a smaller nondegenerate rectangle inside that neighborhood on which the density is bounded below by half its value at $(1,0)$. The rectangle has positive ${\lambda }_B\times {\lambda }_B$ measure by the one-dimensional box formula and the product rectangle formula [F5, F9], hence it has positive probability under every $\theta >0$. Therefore $g(T)$ is not zero almost surely, contradicting completeness [F3].