Does there always exist a function $ f $ for which $ Y - f ( X ) $ and $ X $ are independent?

Question

Let $ X $ and $ Y $ be real random variables.

Does there always exist a function $ f $ for which $ Y - f ( X ) $ and $ X $ are independent?

I tried to prove the statement, but I couldn't do it.

If the statement is false, there must exist random variables $ X $ and $ Y $ such that for any function $ f $, $ Y - f ( X ) $ and $ X $ are not independent.

But I also couldn't find such a pair of random variables $ X $ and $ Y $.

I would appreciate any advise or hint!

Answer 1

No, but there does exist an $f(X)$ such that they are uncorrelated.

Two variables $X$ and $Y$ are independent if the probability distribution of $Y|X$ does not depend on $X$. Consider $Y|X \sim N(0, X^{2})$, then $Y-f(X)|X \sim N(-f(X), X^{2})$ which still depends on $X$ for any function $f$.

If we define $E[f(X)]$ so that $Cov(f(X), X) = Cov(Y,X)$, then $Cov(Y-f(X), X) = 0$. For example, let $f(X) = \frac{Cov(Y,X)}{Var(X)} X$ be linear.

Answer 2

Let $\Omega = \{a,b,c\}$ be a probability space with three outcomes, each having probability $1/3$. Let $X = 1_{\{a\}}$ and $Y = 1_{\{b\}}$. You can check that if $A,B$ are independent events in this space, then one of them must have probability 0 or 1; as a result, any random variable independent of $X$ must be constant. But $Y-f(X)$ can never be constant, as it will necessarily take different values at $b$ and $c$.