Cramer-Rao Lower Bound for the estimation of Pearson correlation Given a bivariate Gaussian distribution $\mathcal{N}\left(0,\begin{pmatrix}
   1 & \rho \\
   \rho & 1 
\end{pmatrix}\right)$, I am looking for information on the distribution of $\hat{\rho}$ when estimating $\rho$ on finite sample with the Pearson estimator.
Is there any known Cramer-Rao lower bound for that?
 A: Yes, there is and it can be derived routinely. The Fisher Information can be shown to be
$$I(\rho) = \frac{1+\rho^2}{\left(1-\rho^2\right)^2}$$
and you know how to get the CRLB from here. The result may be arrived at simply by applying the definition of Fisher Information, i.e. start from the log likelihood
$$\log\left[ f(x;\rho) \right]  =\log\left\{ \frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left\{-\frac{1}{2(1-\rho^2)} \left(x^2 + y^2 - 2\rho xy \right) \right\} \right\}$$
take the derivatives and evaluate the expectation using the properties of the normal distribution. I advise you to verify it on your own.
A: I did the computations on my own, but I find something different:
We consider the set of $2 \times 2$ correlation matrices 
$C = 
\begin{pmatrix}
   1 & \theta \\
   \theta & 1 
\end{pmatrix}$ parameterized by $\theta$.
Let $x = \begin{pmatrix}
   x_1 \\
   x_2 
\end{pmatrix} \in \mathbf{R}^2$.
$f(x;\theta) = \frac{1}{2\pi \sqrt{1-\theta^2}} \exp\left(-\frac{1}{2}x^\top C^{-1} x \right) = 
\frac{1}{2\pi \sqrt{1-\theta^2}} \exp\left( -\frac{1}{2(1-\theta^2)}(x_1^2 + x_2^2 - 2\theta x_1 x_2) \right)$
$\log f(x;\theta) = - \log(2\pi \sqrt{1-\theta^2}) -\frac{1}{2(1-\theta^2)}(x_1^2 + x_2^2 - 2\theta x_1 x_2) $
$\frac{\partial^2 \log f(x;\theta)}{\partial \theta^2} = 
-\frac{\theta^2 + 1}{(\theta^2 - 1)^2}
- \frac{x_1^2}{2(\theta+1)^3}
+ \frac{x_1^2}{2(\theta-1)^3}
- \frac{x_2^2}{2(\theta+1)^3}
+ \frac{x_2^2}{2(\theta-1)^3}
- \frac{x_1 x_2}{(\theta+1)^3}
- \frac{x_1 x_2}{(\theta-1)^3}
$
Then, we compute $\int_{-\infty}^{\infty} \frac{\partial^2 \log f(x;\theta)}{\partial \theta^2} f(x;\theta) dx$.
Since $\mathbf{E}[x_1] = \mathbf{E}[x_2] = 0$, $\mathbf{E}[x_1x_2] = \theta$, $\mathbf{E}[x_1^2] = \mathbf{E}[x_2^2] = 1$, we get
$\int_{-\infty}^{\infty} \frac{\partial^2 \log f(x;\theta)}{\partial \theta^2} f(x;\theta) dx =
- \frac{\theta^2 + 1}{(\theta^2 - 1)^2}
- \frac{1}{2(\theta+1)^3}
+ \frac{1}{2(\theta-1)^3}
- \frac{1}{2(\theta+1)^3}
+ \frac{1}{2(\theta-1)^3}
- \frac{\theta}{(\theta+1)^3}
- \frac{\theta}{(\theta-1)^3}
= 
- \frac{3(\theta^2+1)}{(\theta-1)^2(\theta+1)^2}
$
Thus, $$g(\theta) = \frac{3(\theta^2+1)}{(\theta-1)^2(\theta+1)^2}.$$
