# Foundational sufficient statistics

I've been reading through Casella and Berger's Statistical Infererence and have am having a little trouble understanding something in their explanation of sufficient statistics. Here is the passage from page 272-273 before I proceed (note that $$\boldsymbol X= (X_1,X_2,...,X_n)$$): It could be that I'm overworked, but it has not been too clear to me why $$\{\boldsymbol X=\boldsymbol x\},\{ \boldsymbol Y= \boldsymbol x\} \subseteq \{T(\boldsymbol X)=T(\boldsymbol x)\}$$. Is it the case that $$\{T(\boldsymbol X)=T(\boldsymbol x)\}=\{\boldsymbol X: T(\boldsymbol X)=T(\boldsymbol x)\}???$$ I was hoping someone could help clear this up for me.

It's just a matter of understanding the notation.

### Preliminaries

A random variable, such as $$\boldsymbol X$$, is a measurable function on a probability space,

$$\boldsymbol X: \Omega \to \mathbb{R}^n.$$

A statistic $$T$$ is a measurable function

$$T: \mathbb{R}^n \to \mathbb{R}.$$

The composite function

$$T\circ \boldsymbol{X}:\Omega \to \mathbb{R};\ T(\boldsymbol{X})(\omega) = T(\boldsymbol{X}(\omega))$$

is therefore also a random variable.

Let $$\boldsymbol{x}\in\mathbb{R}^n$$: it is a possible value of $$\boldsymbol X$$. Therefore $$t = T(\boldsymbol{x})\in \mathbb R$$ is a possible value of the statistic $$T$$.

### Notation

The set-building notation used in this context is a shorthand--some would say an abuse of--the more explicit mathematical notation

$$\{T(\boldsymbol X) = T(\boldsymbol x)\} = \{\omega\in\Omega\,|\,T(\boldsymbol{X}(\omega)) = T(\boldsymbol{x})\}= \{\omega\in\Omega\,|\,T(\boldsymbol{X}(\omega)) = t\}.$$

Let's call this set $$\mathcal T$$. The mathematical notation clearly exhibits $$\mathcal T$$ as a subset of $$\Omega$$ and, because $$T\circ \boldsymbol X$$ is measurable, it is an event. It is the set of all outcomes where the value of $$T\circ\boldsymbol{X}$$ equals a given value of the statistic $$T$$, namely $$T(\boldsymbol{x}) = t$$. In other words, $$\mathcal T$$ consists of all outcomes where the statistic $$T$$ has the value $$t$$.

Similarly, the other set-building notations used in the quotation should be interpreted as

$$\boldsymbol{X}^{*}(\boldsymbol{x}) = \{\boldsymbol{X} = \boldsymbol{x}\} = \{\omega\in\Omega\,|\,\boldsymbol{X}(\omega) = \boldsymbol{x}\}$$

$$\boldsymbol{Y}^{*}(\boldsymbol{x}) =\{\boldsymbol{Y} = \boldsymbol{x}\} = \{\omega\in\Omega\,|\,\boldsymbol{Y}(\omega) = \boldsymbol{x}\}.$$

Both of these are events: $$\boldsymbol{X}^{*}(\boldsymbol{x})$$ is the event where the value of $$\boldsymbol X$$ is $$\boldsymbol x$$ and $$\boldsymbol{Y}^{*}(\boldsymbol{x})$$ is the event where the value of $$\boldsymbol Y$$ is $$\boldsymbol x$$. They needn't be the same event, because $$X$$ and $$Y$$ are not necessarily the same random variable.

### Solution

By construction, if we apply $$T\circ \boldsymbol{X}$$ to any element $$\omega$$ of either of these sets, we will obtain the value $$T(\boldsymbol{x}) = t$$, which by definition means $$\omega\in\mathcal T$$. We have merely observed that

$$\boldsymbol{X}(\omega) = \boldsymbol{x} \implies T(\boldsymbol{X}(\omega)) = t$$

and

$$\boldsymbol{Y}(\omega) = \boldsymbol{x} \implies T(\boldsymbol{Y}(\omega)) = t,$$

immediately proving that

$$\boldsymbol{X}^{*}(\boldsymbol{x}) \subset \mathcal T$$

and

$$\boldsymbol{Y}^{*}(\boldsymbol{x}) \subset \mathcal T,$$

QED.

• This was incredibly concise, whuber. I understand fully now. Thank you so much! Once I get to +15 reputation, expect an up-vote. May 7 '15 at 16:31