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Suppose I have a large population of data points $(x,y)$ and that the Pearson's correlation is

$$\textrm{corr}(X,Y) = \rho$$

What can I reasonably say about the correlation I expect to observe in a sample of size $n$? If the sample correlation is $\rho_s$, roughly what is the spread is $\rho_s$? Is $\rho_s$ biased?

If we make some assumptions like normality, can we compute the exact likelihood function of $\rho_s$ as a function of $\rho$?

(Ultimately, I am wondering about the problem of whether an observed high correlation is a fluke or not, and all I have is the sample size and the correlation.)

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onestop provided an answer that will hopefully give you enough to go on. If you really want to know about the distribution of the sample correlation coefficient itself, then the definite reference is: Hotelling, H. (1953). New light on the correlation coefficient and its transforms. Journal of the Royal Statistical Society, Series B, 15, 193-232. Note that this isn't light reading. –  Wolfgang Feb 15 '11 at 10:30
    
I don't think your graphs are right. I've just drawn some graphs of the distribution derived from the Fisher formula which show it is correctly centered. In fact it's pretty obvious from the formula that it must be asympototically unbiased for $N \rightarrow \infty$. Could you post the mathematical core of your code? –  onestop Feb 16 '11 at 9:51
    
@onestop Sure. Added Mathematica code. –  Mark Eichenlaub Feb 16 '11 at 10:04
    
That's not how pdfs transform -- it's a bit more complicated. See en.wikipedia.org/wiki/… –  onestop Feb 16 '11 at 10:11
    
@onestop Of course. Thank you. I realized there was a problem after I posted the code, but it would have taken me a while to figure out how to fix it. –  Mark Eichenlaub Feb 16 '11 at 10:29
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1 Answer

up vote 7 down vote accepted

To quote the Wikipedia article on the Fisher transformation :

If $(X, Y)$ has a bivariate normal distribution, and if the $(X_i, Y_i)$ pairs used to form the sample correlation coefficient $r$ are independent for $i = 1, \ldots, n,$ then $$z = {1 \over 2}\ln{1+r \over 1-r} = \operatorname{arctanh}(r)$$ is approximately normally distributed with mean ${1 \over 2}\ln{{1+\rho} \over {1-\rho}},$ and standard error ${1 \over \sqrt{N-3}},$ where $N$ is the sample size.

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sorry to un-accept. When I tried to use this answer, I found it did not work for the situation I'm interested in (high correlation coefficients). –  Mark Eichenlaub Feb 16 '11 at 9:06
    
@Mark, I did some simulations with R, everything holds pretty good for correlation 0.75 –  mpiktas Feb 16 '11 at 10:27
    
@mpiktas Yes, you're right thank you. I made a mistake in my notebook. –  Mark Eichenlaub Feb 16 '11 at 10:28
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