2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ The range is nowhere near being an ancillary statistic, since its probability distribution depends very much on $\theta. \qquad$ $\endgroup$ – Michael Hardy Apr 9 at 16:29
1
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ A common approach is to look at the derivative in $\theta$ of $\operatorname{E}_{\theta}(g(T))$. $\endgroup$ – Xi'an Mar 13 at 11:07
  • $\begingroup$ Yes, I would have mentioned that if the OP had sought some clarification. $\endgroup$ – StubbornAtom Mar 13 at 11:11
  • 1
    $\begingroup$ @user46697 That is not what I wrote. Observe that $-t<X_1,\ldots,X_n<2t \iff \max \left(-X_{(1)}, \frac{X_{(n)}}{2} \right)<t$. It is by the same argument you found the sufficient statistic $T$ in the first place. $\endgroup$ – StubbornAtom Mar 14 at 17:52
  • 1
    $\begingroup$ @user46697 It is not max of $X_1,\ldots,X_n$; it is max of $-X_{(1)}$ and $X_{(n)}/2$. Had it been the former, it would be correct that all observations are $<t$. $\endgroup$ – StubbornAtom Mar 14 at 18:19
  • 1
    $\begingroup$ That would become apparent once you write down $E_{\theta}(g(T))$ explicitly, realising that it is a function of $\theta$. So in order to prove/disprove completeness of $T$, you can differentiate both sides of the equation $E_{\theta}(g(T))=0$ w.r.t $\theta$ and try to show whether $g(T)$ is identically zero or not. $\endgroup$ – StubbornAtom Mar 14 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.