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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)

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When asking yourself a question of this kind, you might want to test it first on any example in your collection of favorite simple examples. This is what Chris and Henning did, you might also have tested the case when X and Y are independent Bernoulli and f is identity. –  Did Oct 5 '11 at 14:47
    
You're right, I'll take this advice to heart. –  Johannes L Oct 5 '11 at 15:15
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Not even $X+Y$ and $X$ are independent in general. We can see this quickly at the informal level. For example, let $Y$ be always fairly close to $0$, and let $X$ be big with non-zero probability, and close to $0$ otherwise. If $X+Y$ is big, we know $X$ must be. So information about $X+Y$ tells us a lot about $X$. –  André Nicolas Oct 5 '11 at 15:25

3 Answers 3

up vote 2 down vote accepted

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).

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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

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

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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.

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