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(I went through some related StackExchange questions but none addresses the one I have. If you do find one that addresses mine below, I'd be happy to look into that.)

In Chapter 6.2 of Casella and Berger's Statistical Inference where the topic of sufficient statistics is discussed, I couldn't understand a statement that has been mentioned a few times:

... Note that events {$\mathbf{X} = \mathbf{x}$} ... subsets of {$\mathit{T}(\mathbf{X}) = \mathit{T}(\mathbf{x})$}.

I understood (I think) that, with a sample space $x \in \mathcal{X}$, a statistic (transformation function) $\mathit{T}$, and the image $T(x) \in \mathcal{T}$, there is a partition of elements in $\mathcal{X}$ that corresponds to an element in $\mathcal{T}$. What I don't understand is, how is {$\mathbf{X} = \mathbf{x}$} a subset of {$T(\mathbf{X}) = T(\mathbf{x})$} if they're not in the same space?

Khai Yi Chin
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    For any function $f$ you have that if $x=y$ then $f(x)=f(y)$. A statistic is just a special type of function. Hence if $X=x$ then $T(X)=T(x)$, meaning ${X=x}\subseteq {T(X)=T(x)}$. Both sets are subsets of the sample space. – Small Deviation Jan 29 '23 at 16:50
  • What do you mean by that they are not in the same space? For fixed random vector $X$, the sets ${X = x}$ and ${T(X) = T(x)}$ are both measurable subsets of $\Omega$ – Esgeriath Jan 29 '23 at 17:05
  • @Esgeriath my understanding was $\mathcal{X}$ is the domain and $\mathcal{T}$ is the image, so the elements of the two spaces are separated.

    For example, if I have $X = (X_1, X_2) \sim Bernoulli(p)$, then $\mathcal{X} = {(0,0), (0,1), (1,0), (1,1) }$. Suppose the statistic is $\sum X_i$, this means $\mathcal{T} = {0, 1, 2}$. This is kind of how I understood it when I mentioned that they aren't in the same space. Clearly, there's some flaw in my understanding (perhaps about some assumptions that I unknowingly made).

     – Khai Yi Chin Jan 30 '23 at 20:11
  • @KhaiYiChin this is standard convention in probability and statistics: ${X = x} := {\omega \in \Omega : X(\omega) = x}$. In your case, I think you restrict situations only to values taken by $X$, so $\mathcal X = X(\Omega)$, and $\mathcal T = T(\mathcal X)$. But the sets ${X = x}$ and ${T(X) = T(x)}$ are still subsets of $\Omega$, not $\mathcal X$ or $\mathcal T$. I'm more of a probability person, so I might be wrong about some stats conventions though – Esgeriath Jan 30 '23 at 20:19

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