Derivation of the standard error for Pearson's correlation coefficient I am wondering how to derive the formula for the standard error of Pearson's correlation coefficient which is given in Zar for example as
$$
\newcommand{\cov}{{\rm Cov}}
\newcommand{\var}{{\rm Var}}
\newcommand{\sd}{{\rm SD}}
SE_r =\sqrt{\frac{1-r^2}{n-2}}$$
I tried to get it from estimating the variance of r when  
$$r =\frac{\cov(x,y)}{\sd(x)\sd(y)}$$
and $V(X) = E(X^2) - E(X)^2$ so we get $Var(r) = E\bigg(\frac{\cov(x,y)^2}{\var(x)\var(y)}\bigg) - r^2$. But from here I don't know how to continue since $E\bigg(\frac{\cov(x,y)^2}{\var(x)\var(y)}\bigg)$ would have to be $\frac{1-(n-3)r^2}{n-2}$ to get finally to
$$\var(r) =\frac{1-r^2}{n-2}$$
Any suggestions or references where I could look this up?
 A: After looking for a long time for an answer to this same question, I found a couple interesting links:
$\bullet$ The Standard Deviation of the Correlation Coefficient, where we can only see the first page but that's where the derivation is. The "standard deviation by dr Sheppard" is given by something called the Asymptotic distribution of moments, of which you can see a bit in the following source.
$\bullet$ A History of Parametric Statistical Inference from Bernoulli to Fisher, 1713-1935.
The reason for the "n-2" instead of "n" in the root, is that your formula assumes a t-distribution with n-2 degrees of freedom, while the one in the links assumes a normal distribution.
A: I do not have the answer, but for me there is an error in the formula of the question.
It is: $$SE_r =\frac{1-r^2}{\sqrt{n-2}}$$ and 
$$\newcommand{\Var}{\operatorname{Var}} \Var(r)=\frac{(1-r^2)^2}{n-2}$$
I will try to check this by simulation:
library(MASS)

N = 100000
r = 0.8
n = 100

Sigma = matrix(c(1, r, r, 1), nrow=2)
r_obs =  replicate(N, cor(mvrnorm(n, c(0,0), Sigma))[2,1])

> mean(r_obs)
[1] 0.7984783
> sd(r_obs)
[1] 0.03690896

So the standard error for r=0.8 with n=100 is approximately 0.037.
If I use the formula of the question I get:
> sqrt((1-r^2)/(n-2))
[1] 0.06060915

And with the formula I gave:
> (1-r^2)/sqrt((n-2))
[1] 0.03636549

The second formula seems to be much closer to the true value than the first.
Edit:
To try to explain why there are two different formulas for the standard error which are circulating, I found that it depends on how you compute it.
In my first simulation, I used Pearson formula to compute the correlation, but one can also use the least square regression coefficient. I can confirm that the latter method has the standard error proposed in the question:
r_obs <- replicate(N, {
    M<-mvrnorm(n, c(0,0), Sigma)
    c(pearson=cor(M, method="pearson")[2,1],
      regression=lm(M[,2]~M[,1])$coef[[2]])
    })

> apply(r_obs, 1, mean)
   pearson regression 
 0.7981580  0.7998433 
> apply(r_obs, 1, sd)
   pearson regression 
0.03707184 0.06094964 

These are two estimators of the correlation which do not have the same variance.
Edit 2:
That try to reconcile the two formulas did not work, because I forgot to normalize the regression coefficient. The formula to compute the correlation from the regression coefficient is:
$$r=\frac{SD(x)}{SD(y)}b$$ with $$E(y|x)=a+b\cdot x$$
By redoing the correct computation, I obtain in fact the same result:
r_obs <- replicate(N, {
    M<-mvrnorm(n, c(0,0), Sigma)
    c(pearson=cor(M, method="pearson")[2,1],
      regression=lm(M[,2]~M[,1])$coef[[2]]*sd(M[,1])/sd(M[,2]))
    })
> apply(r_obs, 1, sd)
   pearson regression 
0.03676248 0.03676248 
> apply(r_obs, 1, mean)
   pearson regression 
 0.7992615  0.7992615 

Which is reassuring in some way. So I maintain that the standard error formula in the question is incorrect, but maybe I explained where does the error come from.
