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, $X_1$ 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.
 A: I'll add another proof here, the continuous analog of Fang-Yi Yu's proof:
Assume $Y_1$ and $Y_2$ are continuous.  For real numbers $y_1$ and $y_2$, we can define:
$S_{y_1} = \{{x_1: g(x_1)\le y_1} \}$ and
$S_{y_2} = \{{x_2: h(x_2)\le y_2} \}$.
We can then write the joint cumulative distribution function of $Y_1$ and $Y_2$ as:
\begin{eqnarray*}
F_{Y_{1},Y_{2}}(y_{1},y_{2}) & = & P(Y_{1}\le y_{1},Y_{2}\le y_{2})\\
 & = & P(X_{1}\in S_{y_{1}},X_{2}\in S_{y_{2}})\\
 & = & P(X_{1}\in S_{y_{1}})P(X_{2}\in S_{y_{2}})
\end{eqnarray*}
Then the joint probability density function of $Y_{1}$ and $Y_{2}$
is given by:
\begin{eqnarray*}
f_{Y_{1},Y_{2}}(y_{1},y_{2}) & = & \frac{\partial^{2}}{\partial y_{1}\partial y_{2}}F_{Y_{1},Y_{2}}(y_{1},y_{2})\\
 & = & \frac{d}{dy_{1}}P(X_{1}\in S_{y_{1}})\frac{d}{dy_{2}}P(X_{2}\in S_{y_{2}})
\end{eqnarray*}
Since the first factor is a function only of $y_{1}$ and the second
is a function only of $y_{2}$, then we know $Y_{1}$ and $Y_{2}$
are independent (recall that random variables $U$ and $V$ are independent
random variables if and only if there exists functions $g_{U}(u)$
and $h_{V}(v)$ such that for every real $u$ and $v$, $f_{U,V}(u,v)=g_{U}(u)h_{V}(v)$).
A: 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 $f_{2}(X_{2})$.
A: 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 $C_i$.]
A: 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 expand 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)\cdot \sum_{x_2 \in A_2} \Pr(x_2) = \Pr[X_1\in f_1^{-1}(Y_1)]\cdot \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
A: Yes. For two sets $A_1$ and $A_2$, you find
$$P(Y_1\in A_1,Y_2\in A_2)=P(f_1(X_1)\in A_1,f_2(X_2)\in A_2)=P(X_1\in f_1^{-1}(A_1),X_2\in f_2^{-1}(A_2))$$
and
$$P(Y_1\in A_1)P(Y_2\in A_2)=P(f_1(X_1)\in A_1)P(f_2(X_2)\in A_2)=P(X_1\in f_1^{-1}(A_1))P(X_2\in f_2^{-1}(A_2))$$
Since $X_1$ and $X_2$ are independent, it holds
$$P(X_1\in f_1^{-1}(A_1),X_2\in f_2^{-1}(A_2))=P(X_1\in f_1^{-1}(A_1))P(X_2\in f_2^{-1}(A_2))$$
and therefore
$$P(Y_1\in A_1,Y_2\in A_2)=P(Y_1\in A_1)P(Y_2\in A_2)$$
