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Suppose $X_1,..,X_n$ are independent random variables and $X_i'$ is an independent copy of $X_i$, then how does one show that $$E[f(X_1,..,X_n)|X_1,..,X_{i-1},X_{i+1},..,X_n]$$ and $$E[f(X_1,..,X_i',..,X_n)|X_1,..,X_{i-1},X_{i+1},..,X_n]$$ are equal almost everywhere and why is $f(X_1,..,X_n)$ and $f(X_1,..,X_i',..,X_n)$ conditionally independent over $G:=\sigma(X_1,..,X_{i-1},X_{i+1},..,X_n)$?

I am trying to argue using measure theoretic arguments but I can't seem to be able to show this. I am stuck trying to understand the $\sigma$-algebras generated by both functions which are measurable on $\sigma(X_1,..,X_{i-1},X_{i+1},..,X_n)$. Any ideas?

i.e. Possible proof of conditional independence:

Note: $\sigma(f(X_1,..,X_n))\subset \sigma(X_1,..,X_n)$ and $\sigma(f(X_1,..,X_i',..,X_n))\subset \sigma(X_1,..,X_i',..,X_n)$. Let $A\in \sigma(X_1)\cup...\cup\sigma(X_n) $ and $B\in \sigma(X_1)\cup..\cup\sigma(X_i')..\cup\sigma(X_n)$ then $$E[1_A1_B|G]=E[1_A|G]E[1_B|G]$$ This implies that it holds for all $A\in \sigma(X_1,..,X_n)$ and $B\in \sigma(X_1,..,X_i',..,X_n)$.

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    $\begingroup$ I applaud you exploring this question from the fundamental level. Intuitive yes, yet dig in and demonstrate. $\endgroup$ Feb 24 '21 at 4:19
  • $\begingroup$ Uh, why are these independent? Did you mean to write X_i and X'_i in conditioning? (Otherwise take f(x_1, ..., x_n) = x_1 and i=2). (I guess conversely, you might have meant that these are equal almost surely which should be true) $\endgroup$
    – E-A
    Feb 24 '21 at 8:15
  • $\begingroup$ Yes you are right. $f(X_1,..,X_n)$ and $f(X_1,..,X_i',..X_n)$ are conditional independent on $(X_1,..,X_{i-1},X_{i+1},..,X_n)$ $\endgroup$
    – Jhon Doe
    Feb 24 '21 at 8:40
  • $\begingroup$ @E-A Do you know why they might be equal a.e. $\endgroup$
    – Jhon Doe
    Feb 24 '21 at 9:34
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    $\begingroup$ A variable that is independent from Y and has the same distribution. $\endgroup$
    – Jhon Doe
    Feb 24 '21 at 9:50
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Let $Y=[X_1,\ldots,X_{n-1}]$. Then $$ \mathsf{E}[f(X_1,\ldots,X_{n-1},X_n)\mid Y]=\varphi(Y) \quad\text{a.s.} $$ and $$ \mathsf{E}[f(X_1,\ldots, X_{n-1},X_n')\mid Y]=\varphi'(Y) \quad\text{a.s.}, $$ where $\varphi(y)=\mathsf{E}f(y_1,\ldots, y_{n-1}, X_n)$ and $\varphi'(y)=\mathsf{E}f(y_1,\ldots, y_{n-1}, X_n')$. (See, e.g., Lemma 6.2.1 on page 236 here.) But $\varphi(y)=\varphi'(y)$ because $X_n\overset{d}{=}X_n'$.

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  • $\begingroup$ Thanks. but where is y from? Is $y\in Y$ and $y_i\in X_i$ $\endgroup$
    – Jhon Doe
    Feb 24 '21 at 10:07
  • $\begingroup$ $y$ is a vector of numbers, where each $y_i$ represents a realization of $X_i$, $1\le i\le n-1$. $\endgroup$
    – d.k.o.
    Feb 24 '21 at 10:36
  • $\begingroup$ Thank you btw is the proof for conditional independence correct? Also do you have any ideas for the following problem: math.stackexchange.com/questions/4036661/… $\endgroup$
    – Jhon Doe
    Feb 24 '21 at 10:50
  • $\begingroup$ @JhonDoe It is not clear how you choose the sets $A$ and $B$. $\endgroup$
    – d.k.o.
    Feb 24 '21 at 10:56
  • $\begingroup$ Oh for instance A is in the generating set of $\sigma(X_1,...,X_n)$ i.e. $A=X_i^{-1}(B)$ for some Borel set. Similarly for B. $\endgroup$
    – Jhon Doe
    Feb 24 '21 at 11:05

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