Sample correlation is also a MLE estimator On page 599 of this book, the author states (without proving) that for random samples $(X_1, Y_1)$, ..., $(X_n, Y_n)$ from a bivariate normal distribution, the sample correlation coefficient
\begin{align}
r &= \dfrac{\sum_{i=1}^n (X_i - \bar X )(Y_i - \bar Y)}{\sqrt{\sum_{i=1}^n (X_i - \bar X )^2 \sum_{i=1}^n(Y_i - \bar Y)^2}} \\
  & = \dfrac{S_{xy}}{\sqrt{S_{xx} S_{yy}}}
\end{align}
is also the maximum-likelihood estimator of $\rho$, the correlation coefficient.
However I don't know how to prove it. For bivariate distribution $(X, Y)$, we have pdf
$$f(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2} \sigma_x \sigma_y} \exp(-\frac{1}{2(1-\rho^2)}[(\frac{x-\mu_x}{\sigma_x})^2 - 2 \rho (\frac{x-\mu_x}{\sigma_x})(\frac{y-\mu_y}{\sigma_y})+(\frac{y-\mu_y}{\sigma_y})^2]).$$
The log-likelihood would yield
$$\ln L = -n \ln (2\pi \sqrt{1-\rho^2} \sigma_x \sigma_y) + \sum -\frac{1}{2(1-\rho^2)}[...]$$
and taking the partial derivative w.r.t. $\rho$ and setting it to be zero doesn't seem to take me anywhere.
I have found some good resources for special case when $\mu_x = \mu_y = 0$ and $\mathbb{E}[X_i^2] = \mathbb{E}[Y_i^2] = 1$, but I am wondering how to approach the general problem, or if someone can direct me to a great resource I would really appreciate it.
 A: Given two random variables $X$ and $Y$, their correlation coefficient is:
$$\rho_{XY} = \frac{Cov(X,Y)}{\sqrt{Var(X)\cdot Var(Y)}}$$
Where $Cov(X,Y)$ is the covariance of $X$ and $Y$, $Var(X)$ is the variance of $X$, and $Var(Y)$ is the variance of $Y$.
According to your book, the maximum likelihood estimator of $\rho_{XY}$ is:
\begin{align}
r &= \dfrac{\sum_{i=1}^n (X_i - \bar X )(Y_i - \bar Y)}{\sqrt{\sum_{i=1}^n (X_i - \bar X )^2 \sum_{i=1}^n(Y_i - \bar Y)^2}} \\
  & = \dfrac{S_{xy}}{\sqrt{S_{xx} S_{yy}}}
\end{align}
Notice that:
$$\frac{1}{n} \sum_{i=1}^n (X_i - \bar X )(Y_i - \bar Y)$$
is the maximum likelihood estimator of the covariance of $X$ and $Y$. Also, notice that:
$$\frac{1}{n} \sum_{i=1}^n (X_i - \bar X )^2$$
is the (biased) maximum likelihood estimator of the variance of $X$ and:
$$\frac{1}{n} \sum_{i=1}^n(Y_i - \bar Y)^2$$
is the (biased) maximum likelihood estimator of the variance of $Y$.
So, the (biased) maximum likelihood estimator of the correlation coefficient is equal to:
$$r = \frac{\frac{1}{n} \sum_{i=1}^n (X_i - \bar X )(Y_i - \bar Y)}{\sqrt{\frac{1}{n} \sum_{i=1}^n (X_i - \bar X )^2 \cdot \frac{1}{n} \sum_{i=1}^n(Y_i - \bar Y)^2}}$$
The $\frac{1}{n}$ term is multiplied through in the denominator, so:
$$r = \frac{\frac{1}{n} \sum_{i=1}^n (X_i - \bar X )(Y_i - \bar Y)}{\frac{1}{n} \sqrt{\sum_{i=1}^n (X_i - \bar X )^2 \cdot \sum_{i=1}^n(Y_i - \bar Y)^2}} = \frac{\sum_{i=1}^n (X_i - \bar X )(Y_i - \bar Y)}{\sqrt{\sum_{i=1}^n (X_i - \bar X )^2 \cdot \sum_{i=1}^n(Y_i - \bar Y)^2}}$$
Note that this follows from the fact that a function of maximum likelihood estimators is a maximum likelihood estimator.
