Difficulty in showing completeness

Question

Suppose we have the distribution $U(\theta_1-\theta_2,\theta_1+\theta_2)$ ,where $\theta_1 \in \mathbb{R}, \theta_2 >0$. We know that the order statistics$(X_{(1)},X_{(n)})$ are jointly sufficient for $ (\theta_1,\theta_2)$. But I cannot show whether it is complete also. Is there a way to show completeness?

Answer 1

We want to show $$ \forall \theta \hspace{.4cm} E(h(X_{(1)},X_{(n)}))=0 \Rightarrow P(h(X_{(1)},X_{(n)})=0)=1$$

$$E(h(X_{(1)},X_{(n)}))=\iint h(x_1,x_n)f_{X_{(1)},X_{(n)}}(x_1,x_n)=0$$

$$\Rightarrow \int_{\theta_1 - \theta_2}^{\theta_1 + \theta_2} \int_{\theta_1 - \theta_2}^{x_n} h(x_1,x_n)(x_n-x_1)^{n-2}dx_1 dx_n=0$$

define $\mu=\theta_1 - \theta_2$ and $\sigma=\theta_1 + \theta_2$

$$\int_{\mu}^{\sigma} \int_{\mu}^{x_n} h(x_1,x_n)(x_n-x_1)^{n-2}dx_1 dx_n=0$$

by derivation of $\sigma$

$$ \int_{\mu}^{\sigma} h(x_1,\sigma)(\sigma-x_1)^{n-2}dx_1 =0$$

by derivation of $\mu$

$$ - h(\mu,\sigma)(\sigma-\mu)^{n-2}=0 \hspace{.5cm} \forall \mu ,\forall \sigma $$

$$\Rightarrow P(h(X_{(1)},X_{(n)})=0)=1$$