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Let $X_1, \dots, X_n$ be iid. $\text{Uniform}[-\theta,\theta]$. I need to find the complete sufficient statistic for $\theta$ or prove there does not exist such.

I know that $T = (X_{(1)}, X_{(n)} )$ is a sufficient statistic for $\theta$ but it is not a complete sufficient statistic.

I want to prove it. So first I tried to use the Basu's theorem . But in this case $R = X_{(n)} - X_{(1)}, $ is not an ancillary statistic.

So I tried prove using the definition of the complete sufficient statistic.

Here I have attached my work so far: enter image description here

But by doing like this , seems like that I am going to prove that $T$ is a complete sufficient statistic.

So can someone help to figure it out what I did incorrectly ?

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  • $\begingroup$ Since $T$ is a two-dimensional statistic, the expected value will need to be a double integral for arbitrary $g(T)$. $\endgroup$
    – knrumsey
    May 18, 2018 at 0:40
  • $\begingroup$ is it necessary ? because i found the joint distribution of $X_{(1)}$ and $X_{(n)}$. $\endgroup$
    – Sam88
    May 18, 2018 at 0:42
  • $\begingroup$ Yes it is still necessary. By defintion, $E(g(X,Y)) = \int_x \int_y g(x,y)f(x,y)dx \ dy$ where $f(x,y)$ is the joint distribution. That's where you're going wrong. $\endgroup$
    – knrumsey
    May 18, 2018 at 0:44
  • $\begingroup$ Since $\theta$ is a scale parameter, you need to find a function of $(X_{(1)},X_{(n)})$ that is scale free... $\endgroup$
    – Xi'an
    Oct 12, 2018 at 18:07
  • $\begingroup$ For instance, $X_{(1)}/X_{(n)}$! $\endgroup$
    – Xi'an
    Oct 12, 2018 at 18:08

2 Answers 2

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Recall:

Definition: A statistic $T$ is complete for $\theta$ if $$E(g(T)) = 0, \ \text{ for all $\theta$} \quad \Rightarrow \quad P(g(T) = 0) = 1, \ \text{ for all $\theta$}$$

The part about $P(g(T) = 0) = 1$ basically says that the function $g$ is trivially $0$ everywhere (except possibly on a set of measure 0).

So... If you want to prove that $T$ is NOT complete, you can try to find a non-trivial function $g(T)$ for which $E(g(T)) = 0$ for all values of $\theta$.

Hint: Can you find $E(X_{(1)})$ and $E(X_{(n)})$? Start with that, and then try looking at linear combinations of the sufficient order statistics.

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  • $\begingroup$ This is what i exactly tried to do. But somehow i am not getting the desired answer. (please check the solution that i attached). So i am trying to figuring out that. $\endgroup$
    – Sam88
    May 18, 2018 at 0:40
  • $\begingroup$ See my comment above. In this case, you can use $g(T)$ which is just a linear combination of the min and max. This saves you from having to integrate. $\endgroup$
    – knrumsey
    May 18, 2018 at 0:42
  • $\begingroup$ You are suggesting the about finding an ancillary statistic isnt it ? $\endgroup$
    – Sam88
    May 18, 2018 at 0:48
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    $\begingroup$ @knrumsey $E(X_1)=\frac{\theta}{n+1}$ , $E(X_n)=\frac{n\theta}{n+1}$ now how do I proceed I don't get it you said comput these values but how does that help in $E(g(X,Y)) = \int_x \int_y g(x,y)f(x,y)dx \ dy$ how do I use them here ? $\endgroup$
    – Daman
    Oct 12, 2018 at 17:09
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    $\begingroup$ @knrumsey $E((-n)g(x)X_1)+E(g(x)X_n)=0$ This makes it valid to say it's not complete isn't it ? $\endgroup$
    – Daman
    Oct 12, 2018 at 17:22
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Method 1

$(X_{(1)},X_{(n)})$ is not complete because we can find $g\neq0$ but $\mathbb{E}\left[g(X_{(1)},X_{(n)})\right]=0,\forall\theta$. $g$ is $(t_1,t_2)\rightarrow\frac{n+1}{n-1}t_2-\frac{n+1}{1-n}t_1$.

This is because $\mathbb{E}(X_{(n)})=\frac{n-1}{n+1}\theta$ and $\mathbb{E}(X_{(1)})=\frac{1-n}{n+1}\theta$. Thus $\mathbb{E}\left[g(X_{(1)},X_{(n)})\right]=\mathbb{E}\left[\frac{n+1}{n-1}X_{(n)}-\frac{n+1}{1-n}X_{(1)}\right] = \frac{n+1}{n-1}\mathbb{E}(X_{(n)})-\frac{n+1}{1-n}\mathbb{E}(X_{(1)}) = \theta-\theta=0,\forall \theta$.

Method 2

If the sufficient statistic $(X_{(1)},X_{(n)})$ is complete, then it is a minimal sufficient statistic. However, (X_{(1)},X_{(n)}) is not a minimal sufficient statistic. A minimal sufficient statistic is $\max\{-X_{(1)},X_{(n)}\}$. It is possible that $(x_{(1)},x_{(n)})\neq(y_{(1)},y_{(n)})$ but $\max\{-x_{(1)},x_{(n)}\}=\max\{-y_{(1)},y_{(n)}\}$.

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