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 $\boldsymbol X= (X_1,X_2,...,X_n)$):

Sufficent passage

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.


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 $T$.


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.


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

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


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

immediately proving that

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


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


  • 1
    $\begingroup$ This was incredibly concise, whuber. I understand fully now. Thank you so much! Once I get to +15 reputation, expect an up-vote. $\endgroup$
    – user146925
    May 7 '15 at 16:31

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