Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 10 down vote accepted

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 $C_i$.]

share|improve this answer
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

Yes, they are independent.

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 $f_{2}(X_{2})$.

share|improve this answer
Thank you very much. I am studying an introductory course and it seems to be a little hard for me to get things too serious. –  LLS Nov 5 '10 at 11:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.