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RichardKennaway comments on Causality does not imply correlation - Less Wrong
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Comments (54)
Not quite, it's that as a and b go to infinity,
(\int_{a,b}f(x)f'(x)dx)/(b-a))
goes to zero. \int_{a,b}f(x)f'(x)dx = [ f(x)^2/2 ]^b_a, which is bounded, while b-a is unbounded, QED.
LaTeX to Wiki might work, but LaTeX to LW comment doesn't.
Source code:
Formatting tutorial on the Wiki
I tried, but it didn't work for me. I could make a codecogs URL to exhibit the image in my browser, but it got munged when I tried the  embedding.
The problem must be in escape character (see the last section of the wiki article). Try copy-pasting the code I gave above in your comment, and notice the placement of backslashes.
The standard form for correlation coefficient is
cov(x,y)=N(<xy>-<x><y>)
where N is normalisation; it seems that you suppose that <f>=0 and <f'> finite, then. <f>=0 follows from boundedness, but for the derivative it's not clear. If <f'> on (a,b) grows more rapidly than (b-a), anything can happen.
This cannot happen. f is assumed bounded. Therefore the average of f' over the interval [a,b] tends to zero as the bounds go to infinity.
The precise, complete mathematical statement and proof of the theorem does involve some subtlety of argument (consider what happens if f = sin(exp(x))) but the theorem is correct.