The Bivariate Normal Distribution

Two random variables $X$ and $Y$ have a bivariate normal distribution with parameters $(\mu_X, \mu_Y, \sigma_X^2, \sigma_Y^2, \rho)$ if their joint PDF is

$$f_{X,Y}(x,y) = \dfrac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\!\left(-\dfrac{Q(x,y)}{2(1-\rho^2)}\right),$$

where

$$Q(x,y) = \left(\dfrac{x-\mu_X}{\sigma_X}\right)^{\!2} - 2\rho\!\left(\dfrac{x-\mu_X}{\sigma_X}\right)\!\left(\dfrac{y-\mu_Y}{\sigma_Y}\right) + \left(\dfrac{y-\mu_Y}{\sigma_Y}\right)^{\!2}.$$

The parameters $\sigma_X, \sigma_Y > 0$ are the standard deviations and $\rho \in (-1,1)$ is the correlation between $X$ and $Y$. We write $(X,Y) \sim \mathcal{N}_2(\mu_X, \mu_Y, \sigma_X^2, \sigma_Y^2, \rho).$

Equivalently, the distribution is specified by the mean vector and covariance matrix

$$\boldsymbol{\mu} = \begin{bmatrix} \mu_X \\ \mu_Y \end{bmatrix}, \qquad \Sigma = \begin{bmatrix} \sigma_X^2 & \rho\sigma_X\sigma_Y \\ \rho\sigma_X\sigma_Y & \sigma_Y^2 \end{bmatrix}.$$

Notice that $Q(\mu_X, \mu_Y) = 0,$ so the density attains its peak value at the center:

$$f_{X,Y}(\mu_X, \mu_Y) = \dfrac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}.$$