Correlation coefficient: If $\rho = 0$, then $r$ is normally distributed with mean 0. Why? From this source, the estimation of the coefficient of correlation is
$$r = \frac{\Sigma (X_i-E[X])(Y_i-E[Y])}{\sqrt { \Sigma (X_i-E[X])^2 \Sigma (Y_i - E[Y])^2}}$$
If the coefficient correlation is null ($\space\rho = 0\space$), then its estimator $r$ is normally distributed with mean equal 0.
Is this valid whatever is the distribution of $X$ and $Y$? Why?
 A: Oy veh iz mir...
The source is wrong.  In fact, the source acknowledges that it is wrong immediately below the claim when it states that under $H_o:\rho=0$ and when $(X, Y)$ are jointly Normal,
$T = \frac{r}{\sqrt{1-r^2}}\sqrt{n-2} \sim T_{n-2}$ .
Consider the source: the slides are from a BIOLOGY course.  Many biologists are literate statisticians.  Many statisticians are literate biologists.  Very few biologists are experts in statistics, and especially mathematical statistics.  Very few statisticians are experts in biology.  
The proof of the fact (that properly scaled, $R$ follows a $T_{n-2}$ distribution when $\rho = 0$) relies on the underlying Normal.  I'm not sure how distributionally robust that result is.  What is also true (and is pretty distributionally robust) is that 
$Z = \frac{1}{2} \ln \frac{1+r}{1-r} \sim N(\zeta, \frac{1}{n-3})$,
where $\zeta = \tanh^{-1}\rho$.  (Incidentally, $Z = \tanh^{-1} r$.)
A: On slide 2, they assume that $X,Y$ being only normally distributed.
Then if $\rho=0$, which is exactly their correlation parameter, their estimated correlation would be $r=0$ on average naturally, so its mean $0$ aswell.
Why $r$ would also be normally distributed usually requires a longer proof. It may be related to the central limit theorem, that the mean is normally distributed for $n\rightarrow \infty$.
A: One possibility:
You can divide both numerator and denominator by $n$, so the numerator become an average, suggesting we consider the CLT. 
So you might try to apply CLT to that scaled numerator (i.e. the covariance), which application requires some additional manipulation.
If you can apply the CLT to the numerator and invoke Slutsky's theorem, you should be able to show* that asymptotically the sample correlation is normal.
*(if the necessary conditions for both those theorems hold) 
