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It's not hard to prove that, if $X$ and $Y$ are independent random variables then $f(X)$ and $f(Y)$ are independent, where $f,g: \mathbb{R} \rightarrow \mathbb{R}$ are measurable maps (see here) and it's also not hard to extend this definition to random vectors in $\mathbb{R}$. Now for the question:

Suppose we have $X_i$ iid random variables and $Y_i$ iid random variables for $i=1,...,n$, and suppose that mutual independence holds for all $X_i$ and $Y_i$. Now let $X=(X_1,...,X_n)$ and $Y=(Y_1,...,Y_n)$ be random vectors. Since these are independent, can we say that $g(X)=\sum_{i=1}^{n}X_i$ and $g(Y)=\sum_{i=1}^{n}Y_i$ are independent? This seems like a pretty straightforward application, but I have not seen anything like this example referenced in a textbook before, and yet this seems to be a particularly powerful example, so I want to make sure I'm not completely misunderstanding something.

Mr Saltine
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Let $\varepsilon,X_1,X_2,\ldots$ be i.i.d. Rademacher$(\frac12)$ r.v. and let $Y_i:=\varepsilon X_i$. It is easy to check that $X_i$ and $Y_j$ are independent for any $i,j$. However $$g(X):=\sum_{i=1}^nX_i,\quad\text{and}\quad g(Y):=\sum_{i=1}^nY_i=\varepsilon g(X)$$ are not independent (because $|g(X)|=|g(Y)|$ where $|g(X)|$ is non-trivial).

nejimban
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  • So then, what exactly is wrong with what I did in my question if it is indeed true that, if two random vectors $X$ and $Y$ are independent, then $g(X)$ and $g(Y)$ are also independent for measurable $g$? – Mr Saltine Jan 27 '22 at 15:10
  • Might be worth mentioning that the reason why the OP's statement is false because he is only assuming pairwise independence as opposed to mutual independence of all entries. – Andreas Lenz Jan 27 '22 at 15:10
  • @AndreasLenz, I meant to say that X and Y are mutually independent (not pairwise) for all entries. In this case; would what I wrote in my original post hold? – Mr Saltine Jan 27 '22 at 15:13
  • @SimonSaltine: ah, so my answer becomes irrelevant now that you edited your original question! Yes, $\mathbf X:=(X_1,\ldots,X_n)$ and $\mathbf Y:=(Y_1,\ldots,Y_n)$ now become two independent vectors (which was not the case before), so $g(\mathbf X)$ and $g(\mathbf Y)$ are independent as well. – nejimban Jan 27 '22 at 15:16
  • Thank you for clarifying! – Mr Saltine Jan 27 '22 at 15:17
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    @SimonSaltine Just to clarify my comment: Mutual independence of all entries in two random vectors $X$ and $Y$ means $p(x_1,\dots x_n,y_1,\dots,y_n) = p(x_1)\dots p(x_n)p(y_1)\dots p(y_n)$. Independence of two random vectors means $p(x_1,\dots,x_n,y_1,\dots,y_n) = p(x_1,\dots,x_n)p(y_1,\dots,y_n)$. Pairwise independence means $p(x_i,y_j) = p(x_i)p(y_j)$. So mutual independence implies vector independence, which implies pairwise independence. For the OP's property to be true, vector independence is enough. – Andreas Lenz Jan 27 '22 at 15:22