Is f(X+Y), X independent if X,Y are?

Question

Let $X$ and $Y$ be independent random variables and $f$ a smooth function. Is then f(X+Y) and X independent? Or under which conditions? Do you know where I can find something about this?

I know the proposition that $\phi_1(X), \phi_2(Y)$ are then independent for measurable $\phi$ but this doesn't seem to help. In the case of general $h(X,Y)$, the choice $h(X,Y)=X$ would proof it wrong, but this shouldn't be possible for h(X,Y) of the form f(X+Y)

Answer 1

Not necessarily. As a simple counterexample, consider $X$ and $Y$ both uniformly distributed in $[0,1]$ and $f(x)=x$. Then $P(X<0.5 | f(X+Y)>1.5) = 0 \ne P(X<0.5)$

Or even simpler, let $Y$ be a degenerate random value that is always $0$ -- this will be independent of everything, but $X$ and $f(X)$ is far from independent (except if $f$ is also trivial).

Answer 2

No. Consider $X\sim N(0,1)$ and $Y\sim N(0,1)$ and $f=\rm id$. Then

$$\textrm{E}[f(X+Y)X] = \textrm{E}[(X+Y)X] = \textrm{E}[X^2] + \textrm{E}[XY] = 1$$

but

$$\textrm{E}[f(X+Y)] \textrm{E}[X] = \textrm{E}[X+Y] \textrm{E}[X] = 0$$

Answer 3

If $X,Y \sim \operatorname{i.i.d.} N(0,1)$ then $X+Y$ is not independent of $X$, as may be seen by observing that the covariance $\operatorname{cov}(X+Y,X)$ is not zero.