Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Multivariate continuous distributions ▷ Bivariate normal distribution ▷ Construction from standard normal distributions

Theorem: Let $Y_1$ and $Y_2$ be independent random variables following a standard normal distribution:

\[\label{eq:Y12} Y_1 \sim \mathcal{N}(0,1) \quad \text{and} \quad Y_2 \sim \mathcal{N}(0,1) \quad \text{ind.}\]

and define the two random variables $X_1$ and $X_2$

\[\label{eq:X12} \begin{split} X_1 &= \sigma_1 Y_1 + \mu_1 \\ X_2 &= \sigma_2 \left( \rho Y_1 + \sqrt{1-\rho^2} Y_2 \right) + \mu_2 \end{split}\]

where $\mu_1, \mu_2 \in \mathbb{R}$, $\sigma_1, \sigma_2 \in \mathbb{R}_{>0}$ and $\rho \in [-1,+1]$. Then, the random vector with entries $X_1$ and $X_2$ follows a bivariate normal distribution

\[\label{eq:bvn-snorm} \begin{split} X = \left[ \begin{matrix} X_1 \\ X_2 \end{matrix} \right] \sim \mathcal{N}\left( \left[ \begin{matrix} \mu_1 \\ \mu_2 \end{matrix} \right], \left[ \begin{matrix} \sigma_1^2 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma_2^2 \end{matrix} \right] \right) \end{split}\]

where mean and covariance are functions of the parameters $\mu_1$, $\mu_2$, $\sigma_1$, $\sigma_2$ and $\rho$.

Proof: The probability density function of the standard normal distribution is

\[\label{eq:snorm-pdf} Y \sim \mathcal{N}(0,1) \quad \Rightarrow \quad f_Y(y) = \frac{1}{\sqrt{2 \pi}} \cdot \exp \left[ -\frac{1}{2} y^2 \right] \; .\]

The random variables $X_1$ and $X_2$ are functions of $Y_1$ and $Y_2$

\[\label{eq:X12-Y12} \begin{split} X_1 &= \sigma_1 Y_1 + \mu_1 \\ X_2 &= \sigma_2 \left( \rho Y_1 + \sqrt{1-\rho^2} Y_2 \right) + \mu_2 \; , \end{split}\]

such that the inverse functions $Y_1$ and $Y_2$ in terms of $X_1$ and $X_2$ are

\[\label{eq:Y12-X12} \begin{split} Y_1 &= (X_1 - \mu_1)/\sigma_1 \\ Y_2 &= \left[ (X_2-\mu_2)/\sigma_2 - \rho Y_1 \right] / \sqrt{1-\rho^2} \\ &= \left[ (X_2-\mu_2)/\sigma_2 - \rho (X_1-\mu_1)/\sigma_1 \right] / \sqrt{1-\rho^2} \; . \end{split}\]

This implies the following Jacobian matrix

\[\label{eq:Y12-X12-jac} J = \left[ \begin{matrix} \frac{\mathrm{d}Y_1}{\mathrm{d}X_1} & \frac{\mathrm{d}Y_1}{\mathrm{d}X_2} \\ \frac{\mathrm{d}Y_2}{\mathrm{d}X_1} & \frac{\mathrm{d}Y_2}{\mathrm{d}X_2} \end{matrix} \right] = \left[ \begin{matrix} \frac{1}{\sigma_1} & 0 \\ \frac{\rho}{\sigma_1 \sqrt{1-\rho^2}} & \frac{1}{\sigma_2 \sqrt{1-\rho^2}} \end{matrix} \right]\]

and determinant of the Jacobian matrix

\[\label{eq:Y12-X12-jac-det} \begin{split} \lvert J \rvert &= \frac{1}{\sigma_1} \cdot \frac{1}{\sigma_2 \sqrt{1-\rho^2}} - 0 \cdot \frac{\rho}{\sigma_1 \sqrt{1-\rho^2}} \\ &= \frac{1}{\sigma_1 \sigma_2 \sqrt{1-\rho^2}} \\ &= \frac{1}{\sqrt{\sigma_1^2 \sigma_2^2 (1-\rho^2)}} \; . \end{split}\]

Because $Y_1$ and $Y_2$ are independent, the joint density of $Y_1$ and $Y_2$ is equal to the product of the marginal densities:

\[\label{eq:f-Y12-s1} f_{Y_1,Y_2}(y_1,y_2) = f_{Y_1}(y_1) \cdot f_{Y_2}(y_2) \; .\]

Substituting \eqref{eq:snorm-pdf} into \eqref{eq:f-Y12-s1}, we get:

\[\label{eq:f-Y12-s2} \begin{split} f_{Y_1,Y_2}(y_1,y_2) &= \frac{1}{\sqrt{2 \pi}} \exp \left[ -\frac{1}{2} y_1^2 \right] \cdot \frac{1}{\sqrt{2 \pi}} \exp \left[ -\frac{1}{2} y_2^2 \right] \\ &= \frac{1}{\sqrt{(2 \pi)^2}} \cdot \exp \left[ -\frac{1}{2} \left( y_1^2 + y_2^2 \right) \right] \; . \end{split}\]

With the probability density function of an invertible function, the joint density of $X_1$ and $X_2$ can be derived as:

\[\label{eq:f-X12-s1} f_{X_1,X_2}(x_1,x_2) = f_{Y_1,Y_2}(y_1,y_2) \cdot \lvert J \rvert \; .\]

Substituting \eqref{eq:f-Y12-s2}, \eqref{eq:Y12-X12} and \eqref{eq:Y12-X12-jac-det} into \eqref{eq:f-X12-s1}, we get:

\[\label{eq:f-XY12-s2} \begin{split} f_{X_1,X_2}(x_1,x_2) &= \frac{1}{\sqrt{(2 \pi)^2}} \cdot \exp \left[ -\frac{1}{2} \left( \left( \frac{x_1-\mu_1}{\sigma_1} \right)^2 + \left( \left[ \frac{x_2-\mu_2}{\sigma_2} - \rho \frac{x_1-\mu_1}{\sigma_1} \right] / \sqrt{1-\rho^2} \right)^2 \right) \right] \cdot \frac{1}{\sqrt{\sigma_1^2 \sigma_2^2 (1-\rho^2)}} \\ &= \frac{1}{\sqrt{(2 \pi)^2 \sigma_1^2 \sigma_2^2 (1-\rho^2)}} \cdot \exp \left[ -\frac{1}{2} \left( \left( \frac{x_1-\mu_1}{\sigma_1} \right)^2 + \frac{1}{1-\rho^2} \left[ \left( \frac{x_2-\mu_2}{\sigma_2} \right)^2 - 2 \rho \left( \frac{x_1-\mu_1}{\sigma_1} \right) \left( \frac{x_2-\mu_2}{\sigma_2} \right) + \rho^2 \left( \frac{x_1-\mu_1}{\sigma_1} \right)^2 \right] \right) \right] \\ &= \frac{1}{2 \pi \sigma_1 \sigma_2 \sqrt{1-\rho^2}} \cdot \exp \left[ -\frac{1}{2(1-\rho^2)} \left( \left( \frac{x_1-\mu_1}{\sigma_1} \right)^2 - 2 \rho \frac{(x_1-\mu_1)(x_2-\mu_2)}{\sigma_1 \sigma_2} + \left( \frac{x_2-\mu_2}{\sigma_2} \right)^2 \right) \right] \; . \end{split}\]

This is the probability density function of the bivariate normal distribution with parameters

\[\label{eq:bvn-mu-Sigma} \mu = \left[ \begin{matrix} \mu_1 \\ \mu_2 \end{matrix} \right] \quad \text{and} \quad \Sigma = \left[ \begin{matrix} \sigma_1^2 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma_2^2 \end{matrix} \right] \; .\]

Thus, it follows that

\[\label{eq:bvn-snorm-qed} \begin{split} X \sim \mathcal{N}(\mu, \Sigma) \; . \end{split}\]
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Metadata: ID: P502 | shortcut: bvn-snorm | author: JoramSoch | date: 2025-05-22, 15:59.