# 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.

## 1 Answer

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.