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Suppose $X$ and $Y$ are real-valued random variables, and $f$ and $g$ are Borel measurable real-valued functions defined on $\mathbb{R}$.

If $X$ and $Y$ are independent, then I know that $f(X)$ and $g(Y)$ are also independent.

If $X$ and $Y$ are uncorrelated, are $f(X)$ and $g(Y)$ also uncorrelated?

Thanks and regards!

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    No, and the same counterexample from the last thread works.2012-02-11
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    What have you tried? What are your thoughts on this? Have you played around with simple examples? Which ones? What have you discovered?2012-02-11
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    @cardinal: Sorry.2012-02-11
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    @Tim: There is no reason to be sorry. But, here is my point: You ask a lot of questions, many of them interesting and some of them quite insightful. It is obvious you are trying to learn. The *very* best way to learn is to play around and try things on your own. Think about the definitions, what they mean, and why they're there. Use simple examples as toy models to see what makes things work and what makes them break. Practicing this will accelerate your learning considerably.2012-02-12
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    @cardinal: Thanks for the advice! I try my best to keep my questions at high quality, but I can break down from time to time. It is just too hard sometimes. I don't have very solid foundation yet.2012-02-12

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Let $X$ have continuous distribution that is symmetric about the origin, and such that the second moment of $X$ exists. For example, $X$ could have standard normal distribution. Let $Y=|X|$.

Then the correlation coefficient of $X$ and $Y$ is $0$, for by symmetry the mean of each of $X$ and $XY$ is $0$. Let $f(x)=|x|$. The correlation coefficient of $f(X)$ and $f(Y)$ is not $0$.

There are simpler examples. For instance we could let $X$ take on the values $1$ and $-1$, each with probability $1/2$, and let $Y$ and $f$ be as above.

The Story: An Engineering graduate once came to see me. He had done some measurements to study the relationship between fecal coliform counts and the tidal current. He had found quite low correlation, and was puzzled because he was confident there was a relationship. Turned out that like a well-trained engineer, he was looking at the (signed) velocity of the current! I suggested using speed, things worked out, and he was somewhat embarrassed. All very nice, except the reason he came to see me is that he had been my student in the standard Engineering statistics course. The usual warnings I had given there about correlation had no effect.

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    Or let f and g be constant.2012-02-11
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    +1. Thanks! Very nice story! I wish I were in your class. Where do/did you teach?2012-02-12
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    I think your "simpler" example may not work as you intend. In particular, $X$ and $Y$ are *independent*, as are $f(X)$ and $f(Y)$. :)2012-02-13
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    It appears I may not have stated my previous comment very well. My point was two-fold: (**1**) The example where $X$ is uniform on $\{-1,+1\}$ and $Y = |X|$ is actually an example of two *independent* random variables, which is not what the OP was asking about, and (**2**) By independence it must be that $f(X)$ and $f(Y)$ are uncorrelated, which they are in the sense that $\mathbb E f(X) f(Y) = \mathbb E f(X) \mathbb E f(Y)$. Both having zero variance, the *correlation coefficient* would seem to be undefined. It's admittedly a bit of a minor quibble, but one I thought worth mentioning. Cheers.2012-02-13
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    (I like the other example, though, of course.)2012-02-13
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Asking myself the same question I actually got the following implication:

$$cov(f(X),g(Y))=0 \quad \forall f,g \text{ meassurable and bounded} \Rightarrow X,Y \text{are independent} $$

A proof would be: $$ 0=cov(f(X),g(Y))=\mathbb{E}[f(X)g(Y)]-\mathbb{E}[f(X)]\mathbb{E}[g(Y)]\quad \forall f,g \\ \Rightarrow \mathbb{E}[f(X)g(Y)]=\mathbb{E}[f(X)]\mathbb{E}[g(Y)]\quad \forall f,g $$ Which results in: \begin{align} \mathbb{P}(X\in B_1,Y\in B_2)&=\mathbb{E}[1_{B_1}(X)1_{B_2}(Y)]\\ &=\mathbb{E}[1_{B_1}(X)]\mathbb{E}[1_{B_2}(Y)]\\ &=\mathbb{P}(X\in B_1)\mathbb{P}(Y\in B_2) \quad \forall B_1,B_2 \in B(\mathbb{R}) \end{align} Which is the definition of independence. I guess the intuition is, that Correlation is only "independence at first glance" (integration over the entire integral). Focusing on certain parts with (indicator-)functions shows you if they are actually independent or not.