Independence of function of random varibles If there are two independent random variables, $X_1$ and $X_2$, then I define two other random variables $Y_1$ and $Y_2$, where $Y_1 = f_1(X_1,X_2)$ and $Y_2 = f_2(X_1,X_2)$. Can $Y_1$ and $Y_2$ be deemed independent? If there is dependence, what would be the penalty for neglecting this dependence?
Does the inference change if $f_1$ and $f_2$ are Boolean functions?
 A: A very practical example of independent $Y_1$ and $Y_2$ in your notation is the Box–Muller transform. One can prove that for independent $X_1$ and $X_2$ that follow standard uniform distribution (which is on $(0, 1)$ interval), random variables
$$
Y_1 = \sqrt{-2 \ln X_1} \cos \left(2 \pi X_2\right), \\
Y_2 = \sqrt{-2 \ln X_1} \sin \left(2 \pi X_2\right)
$$
are independent and follow standard normal distribution.
But $Y_1$ and $Y_2$ can be dependent as well. For the same $X_1$ and $X_2$ define $Y_1 = X_1 + X_2$ and $Y_2 = X_2$. Then one can check  that
$$
\mathrm{P}\left(Y_1 \geq 1/2 \mid Y_2 \geq 1/2\right) = 
\mathrm{P}\left(X_1 + X_2 \geq 1/2 \mid X_2 \geq 1/2 \right) = 1, \\
 \mathrm{P}\left(Y_1 \geq 1/2 \right) = 1 - \mathrm{P}\left(X_1 + X_2 < 1/2 \right) = \frac{7}{8} < 1.
$$
A good exercise is to think of an example where
$Y_1$ and $Y_2$ are uncorrelated, but still dependent.

EDIT: let $X_1$ and $X_2$ be indicator random variables, taking only $0$ and $1$ values with probability $1/2$. If $Y_1 = X_1 \vee X_2$ and $Y_2 = X_2$, then
$$
\mathrm{P}\left(Y_1 = 1 \mid Y_2 = 1\right) = 
\mathrm{P}\left(X_1 \vee X_2 = 1 \mid X_2 = 1\right) = 1, \\
\mathrm{P}\left(Y_1 = 1\right) = \frac{3}{4},
$$
so $Y_1$ and $Y_2$ are dependent. However, if $Y_1 = X_1 \leftrightarrow X_2$ (so it equals $1$ if $X_1 = X_2$) and $Y_2 = X_2$, one can check that
$$
\mathrm{P}\left(Y_1 = 1 \mid Y_2 = 0\right) = \mathrm{P}\left(Y_1 = 1 \mid Y_2 = 1\right) = \frac{1}{2},
$$
which means that $Y_1$ and $Y_2$ are independent.
A: The answer is: yes, they can be dependent, and the penalty can be arbitrarily severe. Example: $f_1(X_1, X2) = X_1,$ and $f_2(X_1, X_2) = X_1.$
