Show that the maximum of $x_1,...,x_n \sim \mathrm{Uniform}(0,\theta)$ is a sufficient statistic for $\theta$. (From definition)

Question

Problem

Show that the maximum of $x_1,...,x_n \sim \mathrm{Uniform}(0,\theta)$ is a sufficient statistic for $\theta$.

Background

This question has been asked before, but most answers tackle the problem with the Factorization Theorem. I am trying to understand the definition of sufficiency given in Wasserman's All of Statistics, so I've been trying to solve this problem without appealing to theorems.

Here is the definition of a sufficient statistic given by Wasserman:

Denote IID $x_1,...,x_n \sim F$ as $x^n$. Write $x^n \leftrightarrow y^n$ if $f(x^n;\theta) = cf(y^n; \theta)$ for some constant $c$ that might depend on $x^n$ and $y^n$ but not on $\theta$. A statistic $T(x^n)$ is sufficient if $T(x^n) \leftrightarrow T(y^n)$ implies that $x^n \leftrightarrow y^n$.

As I wrote up this question, I think I figured out the problem, but I would greatly appreciate if someone can check my proof.

Attempt

The joint PDF of IID $x_1,...,x_n \sim \mathrm{Uniform(0,\theta)}$ can be written as

$$ \begin{align*} f(x^n;\theta) &= \Pi_{i=1}^n \theta^{-1} \mathbf{1}_{[0,\theta]}(x_i)\\ &= \theta^{-n}\mathbf{1}_{[0,\infty]}(\min x^n)\mathbf{1}_{(-\infty,\theta]}(\max x^n) \end{align*} $$

Note: Given $x^n = x_1,...,x_n$, $T(x^n) = \max(x^n)$ is a single number, so $\max T(x^n) = T(x^n)$. Now suppose that $T(x^n) \leftrightarrow T(y^n)$. Then, by definition,

$$f(T(x^n);\theta) = cf(T(y^n); \theta)$$

for some constant $c$ that might depend on $x^n,y^n$.

This implies

$$ \begin{align*} f(x^n;\theta) &= \theta^{-n}\mathbf{1}_{[0,\infty]}(\min x^n)\mathbf{1}_{(-\infty,\theta]}(\max x^n)\\ &= \theta^{-n}\mathbf{1}_{[0,\infty]}(\min T(x^n))\mathbf{1}_{(-\infty,\theta]}(\max T(x^n))\\ &= f(T(x^n);\theta)\\ &= cf(T(y^n);\theta)\\ &= c\theta^{-n}\mathbf{1}_{[0,\infty]}(\min T(y^n))\mathbf{1}_{(-\infty,\theta]}(\max T(y^n))\\ &= c\theta^{-n}\mathbf{1}_{[0,\infty]}(\min y^n)\mathbf{1}_{(-\infty,\theta]}(\max y^n)\\ &= cf(y^n;\theta) \end{align*} $$

Thus $x^n \leftrightarrow y^n$, and we conclude that $T(x^n)$ is a sufficient statistic for $\theta$.

Edit

My question is more about correctly intepreting Wasserman's definition rather than the specific problem at hand. The problem is just one way to flesh-out my understanding of the definition.

Answer 1

If one deconstructs Larry's (rather awful!) definition

Denote iid $X_1,...,X_n\sim F$ as ${X}^n$. Write $x^n↔y^n$ if $f(x^n;θ)=cf(y^n;θ)$ for some constant $c$ that might depend on $x^n$ and $y^n$ but not on $θ$. A statistic $T(x^n)$ is sufficient if $T(x^n)↔T(y^n)$ implies that $x^n↔y^n$.

• it is important to distinguish random objects like $X^N$ from their realisation, like $x^n$. The relation $x^n↔y^n$ holds between a pair of realisations,
• the density of $X^n$ should be written $f_n(x^n;\theta)$ since the dimension of the entry $x^n\in\mathfrak X^n$ depends on $n$,
• the proportionality equation $f(x^n;θ)=cf(y^n;θ)$ should be written more rigorously as $$f(x^n;θ)=c(x^n,y^n)f(y^n;θ)\quad\forall\,\theta\in\Theta$$or, alternatively, as the fact that the mapping$$\theta\longmapsto f(x^n;θ)\big/ f(y^n;θ)$$is constant,
• the statistic $T(X^n)$ should be indexed by $n$ as well, since it maps $\mathfrak X^n$ to another space $\mathfrak T_n$ that may depend on $n$. (In the Uniform example, it does not since $\mathfrak T_n=\mathbb R_+$ for all $n$'s.)
• $T(x^n)↔T(y^n)$ means that$$f_{T_n}(t;\theta)=c(t,t^\prime)f_{T_n}(t^\prime;\theta)\quad\forall\,\theta\in\Theta$$ when $t=T_n(x^n)$, $t^\prime=T_n(y^n)$ and $f_{T_n}(t;\theta)$ is the density of $T_n(X^n)$, which is not the density of the original random object $X^n$ but its so-called push-forward image by $T_n$,
• the definition thus means that, if $T_n$ is sufficient, then, when the mapping $$\theta\longmapsto f_{T_n}(T_n(x^n);\theta)\big/ f_{T_n}(T_n(y^n);\theta)$$is constant, the mapping $$\theta\longmapsto f(x^n;\theta)\big/ f(y^n;\theta)$$is necessarily constant

When considering the illustrative Uniform case, if $T_n(x^n)=\max_{1\le i\le n}x_i$, the density of $T_n(X^n)$ is $$f_{T_n}(t;\theta)=\frac{nt^{n-1}}{\theta^n}\mathbb I_{(0,\theta)}(t)$$ If $(t,t^\prime)$ is such that $$f_{T_n}(t;\theta)=c(t,t^\prime)f_{T_n}(t^\prime;\theta)\quad\forall\,\theta\in\Theta$$it implies that $$\mathbb I_{(0,\theta)}(t)=\mathbb I_{(0,\theta)}(t^\prime)\quad\forall\,\theta\in\Theta$$hence that $t=t^\prime$. And if $T_n(x^n)=T_n(y^n)$, the mapping $$\theta\longmapsto f(x^n;θ)=\theta^{-n}\prod_{i=1}^n \mathbb I_{(0,\theta)}(x_i) =\theta^{-n}\mathbb I_{\theta\ge T_n(x^n)}=\theta^{-n}\mathbb I_{\theta\ge T_n(y^n)}=f(y^n;\theta)$$is the same for $x^n$ and $y^n$, therefore $x^n↔y^n$ and $T_n(X^n)$ is sufficient for all $n\ge 1$'s.

Answer 2

Extended comment.

I am somewhat confused by what it is you are seeking an answer for, and also by your workings. On the workings:

Assuming that by $f(x^n; \theta)$ you refer to the likelihood function $L(\theta)$, I am unable to understand why you have specified your likelihood function to be

$$L(\theta) = \theta^{-n} \mathbf{1}_{[0, \infty]} (\min x^n) \mathbf{1}_{(-\infty, \theta]}(\max x^n)$$

In particular, at first glance I am not entirely sure if you are specifying the condition under which $L(\theta)$ is nonzero correctly. Further, if it is correct, then it would appear to contain some redundancy in the conditions. That is, I am currently unable to see good reason as to why you would not specify the following instead:

$$L(\theta) = \theta^{-n} \mathbf{1}_{(-\infty, \theta]}(\max x^n)$$

Assuming that you are trying to use the definition to show that $T(x^n) = \max \{x_1, ... x_n\}$ is sufficient, aren't you trying to show that assuming the truth of the statement $T(x^n) \leftrightarrow T(y^n)$ implies $x^n \leftrightarrow y^n$? If so, rather than one long chain of equalities, might it not perhaps be more appropriate to present the structure as:

Assume that [...], that is,

$$\begin{align} \theta^{-n} \mathbf{1}_{(-\infty, \theta]}(\max T(x^n)) = c\theta^{-n} \mathbf{1}_{(-\infty, \theta]}(\max T(y^n)) \end{align}$$

Because [...], the above implies that

$$\theta^{-n} \mathbf{1}_{(-\infty, \theta]}(\max x^n) = c\theta^{-n} \mathbf{1}_{(-\infty, \theta]}(\max y^n)$$

Conclude.

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Now perhaps this might just be a distinction in study, or indeed self-study practices, but all I make of the definition is that it's merely a formal way of saying that for a statistic to be sufficient, it must be the case that retention of this statistic alone allows you to compute the likelihood function without further recourse to rest of the data. And that solely with respect to this purpose, you can effectively bin the rest of the data and still compute the likelihood function. And that the inclusion of $c$ in the definition is just a formal way of expressing the fact that the likelihood function is really an equivalence class of functions.

As an entirely separate self-study point, but perhaps relevant, when LW goes over sufficiency in the publicly available CMU introductory theoretical statistics course here and here, he didn't devote any time on the definition you have quoted from the book, nor any space for it in the lecture notes either; rather goes down the expository route of partitions and conditional distributions.