# Are functions of independent variables also independent?

It's a really simple question. However I didn't see it in books and I tried to find the answer on the web but failed. If I have two independent random variables, X1 and X2, then I define two other random variables Y1 and Y2, where Y1 = f1(X1) and Y2 = f2(X2). Intuitively, Y1 and Y2 should be independent, and I can't find a counter example, but I am not sure. Could anyone tell me whether they are independent? Does it depend on some properties of f1 and f2?

Thank you.

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For any two (measurable) sets $A_i$, $i=1,2$, $Y_i \in A_i$ if and only if $X_i \in B_i$, where $B_i$ are the sets { $s : f_i (s) \in A_i$ }. Hence, since the $X_i$ are independent, ${\rm P}(Y_1 \in A_1 , Y_2 \in A_2) = {\rm P}(Y_1 \in A_1) {\rm P}(Y_2 \in A_2)$. Thus, the $Y_i$ are independent (which is intuitively clear anyway). [We have used here that random variables $Z_i$, $i=1,2$, are independent if and only if ${\rm P}(Z_1 \in C_1 , Z_2 \in C_2) = {\rm P}(Z_1 \in C_1) {\rm P}(Z_2 \in C_2)$ for any two measurable sets $C_i$.]

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I had no idea about the theorem of measurable sets and independence. Anyway, it seems to be a valid proof. (But I have no idea what the measurable sets are) –  LLS Nov 5 '10 at 11:54
On the one hand, my answer also assumes that the functions $f_i$ are measurable. On the other hand, the use of the prefix "measurable" (for sets/functions) may be omitted in an introductory setting. –  Shai Covo Nov 5 '10 at 12:30
If you are studying rigorous probability course with sigma-algebras then you may prove it by noticing that the sigma-algebra generated by $f_{1}(X_{1})$ is smaller than the sigma-algebra generated by $X_{1}$, where $f_{1}$ is borel-measurable function.
If you are studying an introductory course - then just remark that this theorem is consistent with our intuition: if $X_{1}$ does not contain info about $X_{2}$ then $f_{1}(X_{1})$ does not contain info about $f_{2}(X_{2})$.