Checking if a minimal sufficient statistic is complete

Question

Let $X_1, \cdots, X_n$ be iid from a uniform distribution $U[-\theta, 2\theta]$ with $\theta \in \mathbb{R}^+$ unknown. Check if the minimal sufficient statistic of $\theta$ is complete.

I found that$$T(X) = \max \left(-X_{(1)}, \frac{X_{(n)}}{2} \right)$$is minimal sufficient but i am having trouble checking if it's complete.

My attempt: Since uniform is a location distribution, using Basu's theorem, the ancillary statistic would be the range. Since the above minimal statistic is not independent of the ancillary statistic, it is not complete. Am I right?

Answer 1

I think you should stick to the definition of a complete statistic. For that, you need to find the distribution of $T$.

For all $0<t<\theta$, the distribution function of $T$ is

\begin{align} P_{\theta}(T\le t)&=P_{\theta}(-t\le X_1,X_2,\ldots,X_n\le 2t) \\&=\left[P_{\theta}(-t<X_1<2t)\right]^n \\&=\left(\frac{t}{\theta}\right)^n \end{align}

So $T$ has pdf

$$f_T(t)=\frac{nt^{n-1}}{\theta^n}\mathbf1_{0<t<\theta}$$

In other words, $T$ is distributed exactly as $\max_{1\le i\le n} Y_i$ where $Y_i$'s are i.i.d $U(0,\theta)$ variables.

That $T$ is a complete statistic is a well-known fact, proved in detail here.

Answer 2

Basu's theorem states that

Let $(P_\theta; \theta \in \Theta)$ be a family of distributions on a measurable space $(X, \mathcal{A})$ and $T,A$ measurable maps from $(X, \mathcal{A})$ to some measurable space $(Y, \mathcal{B})$. If $T$ is a boundedly complete sufficient statistic for $\theta$, and $A$ is ancillary to $\theta$, then $T$ is independent of $A$.

If indeed there exists an ancillary statistic that is not independent from the minimal sufficient statistic, then it would prove that the minimal sufficient statistic is not complete. But the range is not ancillary since it scales in $\theta$ (which is a scale parameter rather than a location parameter).