Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Multivariate continuous distributions ▷ Multivariate normal distribution ▷ Conditional correlation

Theorem: Let $X$, $Y$ and $Z$ be random variables jointly following a multivariate normal distribution:

\[\label{eq:V} V = \left[ \begin{matrix} X \\ Y \\ Z \end{matrix} \right] \sim \mathcal{N}(\mu, \Sigma) \; .\]

Then, the conditional correlation of $X$ and $Y$ given $Z$ can be computed as

\[\label{eq:mvn-corr-cond} \mathrm{Corr}(X,Y|Z) = \frac{\rho_{XY} - \rho_{XZ} \rho_{YZ}}{\sqrt{1-\rho_{XZ}^2} \sqrt{1-\rho_{YZ}^2}}\]

where $\rho_{XY}$, $\rho_{XZ}$ and $\rho_{YZ}$ are the pairwise correlations of the random variables $X$, $Y$ and $Z$.

Proof: In the first part of the proof, we will justify that we can work with a simplified form for $\Sigma$. In the second part of the proof, we will compute the conditional correlation, as implied by its definition, from $\Sigma$.

1) We will denote the entries of $\Sigma$ as follows:

\[\label{eq:Sigma} \Sigma = \left[ \begin{matrix} \sigma_X^2 & \sigma_{XY} & \sigma_{XZ} \\ \sigma_{YX} & \sigma_Y^2 & \sigma_{YZ} \\ \sigma_{ZX} & \sigma_{ZY} & \sigma_Z^2 \end{matrix} \right]\]

where $\sigma_X^2$, $\sigma_Y^2$ and $\sigma_Z^2$ are the variances and $\sigma_{XY}$, $\sigma_{XZ}$ and $\sigma_{YZ}$ are the covariances of the entries of the random vector $V$. Define the following matrix:

\[\label{eq:A} A = \left[ \begin{matrix} \frac{1}{\sigma_X} & 0 & 0 \\ 0 & \frac{1}{\sigma_Y} & 0 \\ 0 & 0 & \frac{1}{\sigma_Z} \end{matrix} \right] \; .\]

With the linear transformation theorem for the multivariate normal distribution, the distribution of $W = AV$ is

\[\label{eq:W} W = AV = \left[ \begin{matrix} X/\sigma_X \\ Y/\sigma_Y \\ Z/\sigma_Z \end{matrix} \right] \sim \mathcal{N}(A \mu, A \Sigma A^\mathrm{T})\]

with

\[\label{eq:A-Sigma-A} A \Sigma A^\mathrm{T} = \left[ \begin{matrix} \frac{\sigma_X^2}{\sigma_X \sigma_X} & \frac{\sigma_{XY}}{\sigma_X \sigma_Y} & \frac{\sigma_{XZ}}{\sigma_X \sigma_Z} \\ \frac{\sigma_{YX}}{\sigma_Y \sigma_X} & \frac{\sigma_Y^2}{\sigma_Y \sigma_Y} & \frac{\sigma_{YZ}}{\sigma_Y \sigma_Z} \\ \frac{\sigma_{ZX}}{\sigma_Z \sigma_X} & \frac{\sigma_{ZY}}{\sigma_Z \sigma_Y} & \frac{\sigma_Z^2}{\sigma_Z \sigma_Z} \end{matrix} \right] = \left[ \begin{matrix} 1 & \rho_{XY} & \rho_{XZ} \\ \rho_{YX} & 1 & \rho_{YZ} \\ \rho_{ZX} & \rho_{ZY} & 1 \end{matrix} \right] \; .\]

Since correlation is invariant under linear transformation, the entries of $V$ and $W$ have the same pairwise correlations and conditional correlations. Thus, for the above theorem, we can assume without loss of generality that

\[\label{eq:Sigma-wlog} \Sigma = \left[ \begin{matrix} 1 & \rho_{XY} & \rho_{XZ} \\ \rho_{YX} & 1 & \rho_{YZ} \\ \rho_{ZX} & \rho_{ZY} & 1 \end{matrix} \right] \; .\]

2) The theorem about conditional distributions of the multivariate normal distribution states that

\[\label{eq:mvn-cond} X \sim \mathcal{N}(\mu, \Sigma) \quad \Rightarrow \quad X_1|X_2 \sim \mathcal{N}(\mu_{1|2}, \Sigma_{1|2})\]

with

\[\label{eq:mvn-cond-para} \begin{split} \mu_{1|2} &= \mu_1 + \Sigma_{12} \Sigma_{22}^{-1} (X_2 - \mu_2) \\ \Sigma_{1|2} &= \Sigma_{11} - \Sigma_{12} \Sigma_{22}^{-1} \Sigma_{21} \end{split}\]

where

\[\label{eq:mvn-joint-para} \mu = \left[ \begin{matrix} \mu_1 \\ \mu_2 \end{matrix} \right] \quad \text{and} \quad \Sigma = \left[ \begin{matrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{matrix} \right] \; .\]

With that, we can derive the conditional distribution of $X$ and $Y$ given $Z$. Specifically, the conditional covariance matrix can be computed as follows:

\[\label{eq:Sigma-cond} \begin{split} \Sigma_{X,Y|Z} &= \Sigma_{XY} - \Sigma_{XY,Z} \Sigma_{Z}^{-1} \Sigma_{Z,XY} \\ &= \left[ \begin{matrix} 1 & \rho_{XY} \\ \rho_{YX} & 1 \end{matrix} \right] - \left[ \begin{matrix} \rho_{XZ} \\ \rho_{YZ} \end{matrix} \right] \left[ \begin{matrix} 1 \end{matrix} \right]^{-1} \left[ \begin{matrix} \rho_{ZX} & \rho_{ZY} \end{matrix} \right] \\ &= \left[ \begin{matrix} 1 & \rho_{XY} \\ \rho_{YX} & 1 \end{matrix} \right] - \left[ \begin{matrix} \rho_{XZ} \rho_{ZX} & \rho_{XZ} \rho_{ZY} \\ \rho_{YZ} \rho_{ZX} & \rho_{YZ} \rho_{ZY} \end{matrix} \right] \\ &= \left[ \begin{matrix} 1 - \rho_{XZ}^2 & \rho_{XY} - \rho_{XZ} \rho_{YZ} \\ \rho_{XY} - \rho_{XZ} \rho_{YZ} & 1 - \rho_{YZ}^2 \end{matrix} \right] = \left[ \begin{matrix} \sigma_{X|Z}^2 & \sigma_{X,Y|Z} \\ \sigma_{X,Y|Z} & \sigma_{Y|Z}^2 \end{matrix} \right] \end{split}\]

With that, we can derive the conditional correlation of $X$ and $Y$ given $Z$. Specifically, dividing the covariance of the conditional distribution by the product of its standard deviations, we get:

\[\label{eq:mvn-corr-cond-qed} \begin{split} \mathrm{Corr}(X,Y|Z) &= \frac{\mathrm{Cov}(X,Y|Z)}{\sqrt{\mathrm{Var}(X|Z)} \sqrt{\mathrm{Var}(Y|Z)}} \\ &= \frac{\sigma_{X,Y|Z}}{\sigma_{X|Z} \sigma_{Y|Z}} \\ &= \frac{\rho_{XY} - \rho_{XZ} \rho_{YZ}}{\sqrt{1-\rho_{XZ}^2} \sqrt{1-\rho_{YZ}^2}} \; . \end{split}\]
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Metadata: ID: P529 | shortcut: mvn-corrcond | author: JoramSoch | date: 2026-03-26, 15:43.