 Solution

A complete statistic $T$ for a family $(P_{\theta })$ if every measurable $g$ with ${\mathbb{E}}_{\theta }|g(T)|<\infty$ and ${\mathbb{E}}_{\theta }g(T)=0$ for all $\theta$ satisfies $g(T)=0$ almost surely under every $P_{\theta }$. It is boundedly complete if the same implication is required only for bounded measurable $g$. Completeness therefore implies bounded completeness.

The vector $X$ is multivariate normal with mean zero and covarianceIts determinant is $(1-\theta {)}^2(1+2\theta )$ and its inverse is $(1-\theta {)}^{-1}I-\theta [(1-\theta )(1+2\theta ){]}^{-1}J$. Hence the joint density isThe Fisher-Neyman factorization theorem gives sufficiency. For minimality use the likelihood-ratio criterion for minimal sufficiency. The density ratio for $x,x^{\prime }$ is independent of $\theta$ precisely whenis constant. Multiplying by $2(1-\theta )(1+2\theta )$ shows that its constant would produce a quadratic term on the right whereas the left is linear, so that constant is zero. The remaining coefficients give $\Delta T_1=\Delta T_2=0$. Thus $T$ is minimal sufficient statistic.

Each $X_i$ is standard normal, so ${\mathbb{E}}_{\theta }{T}_1=3$. The integrable, nonzero function $T_1-3$ has zero expectation for every parameter, proving that $T$ is not complete.

By sufficiency, choose a version $S(T)={\mathbb{P}}_{\theta }(X_1^2\le 1\mid T)$ common to all parameters. The tower property givesOn the event $T_1<1$ we necessarily have $X_1^2<1$, so $S=1$ there almost surely. This event has positive probability under every parameter because the joint normal density is strictly positive on a ball about the origin. Thus $S$ is not identically $c$. The bounded function $S-c$ has zero expectation for every parameter but is not almost surely zero. Consequently