# Independence of functions of independent random variables.

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 $$B\in \sigma(X_1,..,X_i',..,X_n)$$.

• I applaud you exploring this question from the fundamental level. Intuitive yes, yet dig in and demonstrate. Feb 24 '21 at 4:19
• 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)
– E-A
Feb 24 '21 at 8:15
• Yes you are right. $f(X_1,..,X_n)$ and $f(X_1,..,X_i',..X_n)$ are conditional independent on $(X_1,..,X_{i-1},X_{i+1},..,X_n)$ Feb 24 '21 at 8:40
• @E-A Do you know why they might be equal a.e. Feb 24 '21 at 9:34
• A variable that is independent from Y and has the same distribution. Feb 24 '21 at 9:50

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 $$X_n\overset{d}{=}X_n'$$.
• Thanks. but where is y from? Is $y\in Y$ and $y_i\in X_i$ Feb 24 '21 at 10:07
• $y$ is a vector of numbers, where each $y_i$ represents a realization of $X_i$, $1\le i\le n-1$. Feb 24 '21 at 10:36
• @JhonDoe It is not clear how you choose the sets $A$ and $B$. Feb 24 '21 at 10:56
• 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. Feb 24 '21 at 11:05