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As seeing the comment on the 2nd answer of this question, mentioning delta function, I tried to write the proof of that part.

Proof

As $T(\mathbf{x})=t\mathbf{1}_{\mathbf{X}}(\mathbf{x}) \Rightarrow \mathbf{X} = \mathbf{x}$, $\mathbf{X}=\mathbf{x}$ is necessary to $\mathbf{T}(\mathbf{x})=t$.

Therefore, $\{\{\omega \mid\mathbf{X}=\mathbf{x}\} \cap \{\omega \mid\mathbf{T}(\mathbf{x})=t\}\}=\{\omega \mid\mathbf{X}=\mathbf{x}\}$.

$\mathbf{1}_{\mathbf{X}}(\mathbf{x})$ is an indicator function, which is defined

$$ {\displaystyle \mathbf {1} _{X}(x):={\begin{cases}1~&{\text{ if }}~x\in X~,\\0~&{\text{ if }}~x\notin X~.\end{cases}}} $$

Question

Is this correct? Note that I am not used to writing a rigorous proof in English so some notations might be awkward.

  • What are you trying to prove? $P_{\theta}(X=x,T=t)=P_{\theta}(X=x)$ is not true in general unless ${X=x}\subset {T=t}$, in which case this is just a general fact: $$P(A\cap B) = P(A),\quad A\subset B$$, as $A\cap B = A$ – Andrew Jul 15 '23 at 19:09
  • Thanks, @AndrewZhang. I really appreciate your comment. Let me ask you a question. Is it always true that $P_{\theta}(X=x,T=t)=P_{\theta}(X=x)$ if there is an assumption of $T(x) = t$, because $T$ is derives from $X$ and we can say that ${X=x}\subset {T=t}$? That is the only thing that my textbook says about the assumption of this theorem. – CatsEyeblow Jul 15 '23 at 21:26
  • It may make things clearer if you write $T=f(X)$, for some $f$. Then, provided $x\in {u:f(u)=t}$, one has ${ X=x}\subset {T=t}$. – Andrew Jul 15 '23 at 21:52
  • Yes, that is clearer and more obvious. Thank you! – CatsEyeblow Jul 15 '23 at 22:10

1 Answers1

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Please see the comment of the question. Thanks to Andrew, I understand what I was wrong and where to start.

Summary

Given $T(\mathbf{x})=t$, $T$ can be written as $T=f(X)$, for some $f$.

Then for $x\in \{u:f(u)=t\}$, one has $\{ X=x\}\subset \{T=t\}$.

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