I'm doing Exercise 6.26 in Casella and Berger's Statistical Inference, and I'm trying to prove the following:

"Use Theorem 6.6.5 to establish that, given a sample $X_1,...,X_n$, the maximum order statistic $X_{(n)}$ is a minimal sufficient statistic for the uniform$(0,\theta)$ family of distributions."

Theorem 6.6.5 (Minimal sufficient statistics) Suppose that the family of densities $\{f_0(\boldsymbol{X}),...,f_k(\boldsymbol{X})\}$ all have common support. Then

a. The statistic $$T(\boldsymbol{X}) = \bigg( \frac{f_1(\boldsymbol{X})}{f_0(\boldsymbol{X})}, \frac{f_2(\boldsymbol{X})}{f_0(\boldsymbol{X})}, ..., \frac{f_k(\boldsymbol{X})}{f_0(\boldsymbol{X})} \bigg)$$

is minimal sufficient for the family $\{f_0(\boldsymbol{X}),...,f_k(\boldsymbol{X})\}$.

b. If $\mathscr{F}$ is a family of densities with common support, and

(i) $f_i(\boldsymbol{x}) \in \mathscr{F}$, $i=0,1,...,k,$

(ii) $T(\boldsymbol{x})$ is sufficient for $\mathscr{F}$,

then $T(\boldsymbol{x})$ is minimal sufficient for $\mathscr{F}.$

I know how to prove that this statistic is a minimal sufficient statistic for this family of distributions, but I don't know how to prove it this way. Any help would be much appreciated, because I'm getting nowhere with this. Each member of this family of distributions has a different support (namely $(0,\theta)$), so I don't see how this theorem could be applied. But I found a set of notes (https://www.stat.colostate.edu/~riczw/teach/STAT730_S15/Lecture/ST730note.pdf) with the same problem (but no solution), so I'm thinking perhaps this is indeed doable and is not an error in the statement of the problem.

  • $\begingroup$ it seems Theorem 6.6.5 is for a finite family but, $U(0,\theta)$ is not a finite family $\endgroup$
    – Masoud
    Commented Apr 5, 2019 at 1:14
  • $\begingroup$ @masoud, no, $\mathscr{F}$ doesn't need to be finite, and that's where $U(0,\theta)$ should factor in somehow. The family $\{f_0(\boldsymbol{X}),...,f_k(\boldsymbol{X})\}$ is usually a subset of the distribution family you're interested in, and you're looking for some convenient densities whose ratio will give you what you want. It doesn't necessarily have to yield the statistic you want right away, but you're looking for that ratio or those ratios to give you an equivalent one, so that you can then make the necessary conclusion. $\endgroup$
    – Ryker
    Commented Apr 5, 2019 at 13:21
  • $\begingroup$ @Xi'an No, these are different exercises, they just use the same theorem. $\endgroup$
    – Ryker
    Commented Apr 22, 2019 at 20:17
  • 2
    $\begingroup$ I would say that, since the theorem starts with the common support assumption, it clearly does not apply to this case, so the authors may have made a mistake/typo. $\endgroup$
    – Xi'an
    Commented May 8, 2019 at 8:19

1 Answer 1


In uniform distribution, the support of densities varies according to the parameter $\theta$. If we still want to use theorem 6.6.5, we need to assume the support to be all positive numbers and use the indicator function (all positive numbers rather than all real numbers because in the question, we use notation uniform(0,$\theta$) which indicates $\theta$>0).

$$ T(\textbf{X})=(\frac{f_1(\textbf{X})}{f_0(\textbf{X})},\frac{f_2(\textbf{X})}{f_0(\textbf{X})},...) $$

For the first element:

$$ \frac{f_1(\textbf{X})}{f_0(\textbf{X})} = \frac{\theta_1^{-n}\big(I_{(0,\theta_1)}(x_{(n)})\big)}{\theta_0^{-n}\big(I_{(0,\theta_0)}(x_{(n)})\big)}=constant*\frac{I_{(0,\theta_1)}(x_{(n)})}{I_{(0,\theta_0)}(x_{(n)})} $$

The last equality: $\theta_0$ and $\theta_1$ are constants since they are the parameter of a specific element in the density family.

So, $$ T(\textbf{X})= \big( constant_1*\frac{I_{(0,\theta_1)}(x_{(n)})}{I_{(0,\theta_0)}(x_{(n)})}, constant_2*\frac{I_{(0,\theta_2)}(x_{(n)})}{I_{(0,\theta_0)}(x_{(n)})},... \big) $$

And we now prove that this vector $T(\textbf{X})$ and $x_{(n)}$ are one-one correspondance (the following prove is very loose or just by intuition, I am trying to prove a uncountable case with countable set...According to the textbook: Theory of Point Estimation by Lehmann and Casella, 2nd ed, page 38, theory 6.6.5 can be generalized to uncountable case).

Since each element (each $\theta$) in the vector $T(\textbf{X})$ is fixed, we can rearrange them according to the order of parameter $\theta$, i.e.,

$$ T^{*}(\textbf{X})= \big( constant_1*\frac{I_{(0,\theta_{(1)})}(x_{(n)})}{I_{(0,\theta_0)}(x_{(n)})}, constant_2*\frac{I_{(0,\theta_{(2)})}(x_{(n)})}{I_{(0,\theta_0)}(x_{(n)})},... \big) $$

(0) We need to prevent the denominator from being 0. This is possible. A very loose way to do this is set $\theta_0=+\infty$ which ensures the denominator to be 1. Or (read step (2) first), find a $\theta_0\geq\theta_{(3)}$ in the example. Formally, we need $$ \theta_0\geq \min_{\substack{i \\ I_{(0,\theta_{(i)})}(x_{(n)})=1}}\theta_{(i)} $$

(1) If $x_{(n)}$ is known, then there is only one value of $T^{*}(\textbf{X})$ correspond to it.

(2) If the value of $T^{*}(\textbf{X})$ is known, for example $T^{*}(\textbf{X})=(0,0,1,1,1,1,...)$, we know that $\theta_{(2)}<x_{(n)}<\theta_{(3)}$. Since $\theta$ are all real positive numbers and real numbers are dense (uncountable), there is only one $x_{(n)}$ according to this result, i.e., only one $x_{(n)}$ correspond to $T^{*}(\textbf{X})$.

Combiming (1) and (2), $T(\textbf{X})$ and $x_{(n)}$ are one-one correspondance.

According to theorem 6.2.13, it is easy to prove that any 1-1 function of a minimal sufficient statistic is also a minimal sufficient statistic. Thus, $x_{(n)}$ is also a minimal sufficient statistic.


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