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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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    @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

2 Answers 2

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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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    (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.