# asymptotic distribution of correlation coefficient in bivariate normal distribution

Suppose \begin{align} \begin{pmatrix} X_i\\ Y_i \end{pmatrix} \sim_{iid} N \begin{pmatrix} \begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix}, \begin{pmatrix} \sigma^2_1 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma_2^2 \end{pmatrix} \end{pmatrix} \end{align} with $$i=1,...n$$, $$\sigma_1, \sigma_2 >0$$, $$\mu_1,\mu_2 \in \mathbb{R}$$, $$\rho \in (-1,1)$$.

I'm trying to calculate an asymptotic distribution of $$\rho^{mle}$$.

I already know that $$\sqrt{n}(\rho^{mle} - \rho) \rightarrow N(0, (1-\rho^2)^2)$$ but I failed to derive that variance.

To get the fisher information of $$\rho$$, I use

$$log f(X,Y;\rho) = -\frac{1}{2} log(1-\rho^2) - \frac{1}{2(1-\rho^2)}\Big(\big(\frac{X-\mu_1}{\sigma^1}\big)^2 + \big(\frac{Y-\mu_2}{\sigma^2}\big)^2 \Big) + \frac{\rho}{(1-\rho^2)}\Big(\frac{(X-\mu_1)(Y-\mu_2)}{\sigma_1\sigma_2}\Big)$$

to calculate second derivative of that function and then have an expectation.

$$\frac{\partial^2 logf}{\partial \rho^2} = \frac{(1- \rho^4)}{(1-\rho^2)^3} - \frac{1+3\rho^2}{(1-\rho^2)^3}\Big(\big(\frac{X-\mu_1}{\sigma^1}\big)^2 + \big(\frac{Y-\mu_2}{\sigma^2}\big)^2 \Big) + \frac{2\rho(1-\rho^2) + 4\rho(1+\rho^2)}{(1-\rho^2)^3}\Big(\frac{(X-\mu_1)(Y-\mu_2)}{\sigma_1\sigma_2}\Big)$$

When I use that $$E\Big(\big(\frac{X-\mu_1}{\sigma^1}\big)^2 + \big(\frac{Y-\mu_2}{\sigma^2}\big)^2 \Big) = 2$$ and $$E\Big(\frac{(X-\mu_1)(Y-\mu_2)}{\sigma_1\sigma_2}\Big) = \rho$$ to caculate $$-E\big[\frac{\partial^2 logf}{\partial \rho^2}\big] = \frac{1+\rho^2}{(1-\rho^2)^3}$$.

I can't get what I want. Where did I make a mistake?