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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, $X_1$ and $X_2$, then I define two other random variables $Y_1$ and $Y_2$, where $Y_1$ = $f_1(X_1)$ and $Y_2$ = $f_2(X_2)$.

Intuitively, $Y_1$ and $Y_2$ 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 $f_1$ and $f_2$?

Thank you.

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Not seen in the books?? –  Did Nov 16 at 15:16

3 Answers 3

up vote 11 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 $C_i$.]

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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

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

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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

Yes, they are independent.

The previous answers are sufficient and rigorous. On the other hand, it can be restated as followed. Assume they are discrete random variable.

$Pr[Y_1 = f_1(X_1) \wedge Y_2 = f_2(X_2)] = Pr[X_1 \in f_1^{-1}(Y_1)\wedge X_2\in f_2^{-1}(Y_2)] = Pr[X_1 \in A_1 \wedge X_2 \in A_2]$

and we spend it by probability mass function derived

$ = \sum_{x_1 \in A_1\wedge x_2 \in A_2}Pr(x_1, x_2) = \sum_{x_1 \in A_1\wedge x_2 \in A_2}Pr(x_1)Pr(x_2) $

Here we use the independency of $X_1$ and $X_2$, and we shuffle the order of summation

$= \sum_{x_1 \in A_1}Pr(x_1)*\sum_{x_2 \in A_2} Pr(x_2) = Pr[X_1\in f_1^{-1}(Y_1)]*Pr[X_2 \in f_2^{-1}(Y_2)] = Pr[Y_1 = f_1(X_1)]Pr[Y_2 = f_2(X_2)] $

Here we show the function of independent random variable is still independent

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