Let $\vec{X}$ be a random vector following a multi-variate normal distribution $P(\vec X)$ with covariance matrix $\Sigma$ and zero means (for simplicity). Consider a partition of $\vec X$ into two subsets of variables, $\vec X = \{\vec X_1, \vec X_2\}$. What is the mutual information between $\vec X_1$ and $\vec X_2$:

$$I(\vec X_1; \vec X_2) = \int P(\vec X)\ln \frac{P(\vec X)}{P_1(\vec X_1)P_2(\vec X_2)} \mathrm{d}\vec X$$


$$P_1(\vec X_1) = \int P(\vec X)\mathrm{d}\vec X_2, \qquad P_2(\vec X_2) = \int P(\vec X)\mathrm{d}\vec X_1 $$

I presume an analytical answer can be given, but I've been having trouble obtaining it.


2 Answers 2


Actually the answer is trivial to obtain.

$$I(\vec X_1,\vec X_2) = S_1 + S_2 - S_X$$

where $S_1$ is the entropy of $\vec X_1$, $S_2$ the entropy of $\vec X_2$, and $S_X$ the entropy of the full distribution of $\vec X$.

Therefore we obtain the result

$$I(\vec X_1,\vec X_2) = \frac{1}{2} \ln \left( \frac{\det (\Sigma_1) \det (\Sigma_2)}{\det (\Sigma)} \right)$$

where we write the covarinace matrix in the block-form:

$$\Sigma = \left(\begin{array}{cc} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{array}\right)$$

  • $\begingroup$ How can this multivariate form of MI for subsets be extended to its copula entropy equivalent? $\endgroup$
    – develarist
    Dec 10, 2020 at 14:17
  • $\begingroup$ @develarist What do you mean by "its copula entropy equivalent"? Can you expand? Consider asking a new question and notifying me here in the comments. $\endgroup$
    – becko
    Dec 10, 2020 at 14:22
  • $\begingroup$ Ma (2020) showed that $I(\mathbf{X}) = \iint_{\mathbf{u}} c(\mathbf{u}) \ln c(\mathbf{u}) \, d\mathbf{u}$ where $c(\cdot)$ is a multivariate copula density, and $\mathbf{u}$ are the uniform marginals corresponding to $\mathbf{X}$ arxiv.org/abs/2005.14025 $\endgroup$
    – develarist
    Dec 10, 2020 at 14:33

To see where the results comes from using another approach, first recall that if $X,Y$ have joint distribution $p(X,Y)$ and marginals $p(X)$ and $p(Y)$, respectively, then the mutual information is the Kullback-Leibler (KL) divergence between the joint distribution and product of marginal distributions:

$$I(X:Y) = D_{\rm KL}(p(X,Y)\mid\mid p(X)p(Y)).$$

Next, for a pair of $n$-dimensional multivariate normal distributions with mean and covariance parameters $(\mu, \Sigma)$ and $(\tilde{\mu}, \tilde{\Sigma})$, respectively, the KL divergence is given by:

$$ \begin{aligned} &D_{\rm KL}((\mu,\Sigma) \mid\mid (\tilde{\mu}, \tilde{\Sigma})) \\ &= \frac{1}{2} \left[ \log\left( \mathrm{det}(\tilde{\Sigma})/ \mathrm{det}(\Sigma) \right) + \mathrm{tr}\left( \Sigma \tilde{\Sigma}^{-1} \right) - n + (\mu - \tilde{\mu})^\top\tilde{\Sigma}^{-1}(\mu - \tilde{\mu}) \right]. \end{aligned} $$

Suppose now that the joint distribution $p(X,Y)$ is multivariate normal with mean and covariance given by

$$ \mu = \begin{bmatrix} \mu_x \\ \mu_y \end{bmatrix}, \qquad \Sigma = \begin{bmatrix} \Sigma_x & \Sigma' \\ \Sigma' & \Sigma_y \end{bmatrix}. $$

The product of marginals $p(X)p(Y)$ is also multivariate normal, with mean and covariance given by

$$ \tilde{\mu} = \begin{bmatrix} \mu_x \\ \mu_y \end{bmatrix}, \qquad \tilde{\Sigma} = \begin{bmatrix} \Sigma_x & \mathbf{0} \\ \mathbf{0} & \Sigma_y \end{bmatrix}. $$

To complete the computation, we simplify the expression for $D_{\rm KL}$ between two multivariate normal distributions whose are parameters are $(\mu, \Sigma)$ and $(\tilde{\mu}, \tilde{\Sigma})$ as given above, by using the following facts:

  1. Since $\tilde{\Sigma}$ is a block matrix and $\Sigma_y$ is invertible, we have

$$\mathrm{det}(\tilde{\Sigma}) = \mathrm{det}(\Sigma_x - \mathbf{0}\Sigma_y^{-1}\mathbf{0})\mathrm{det}(\Sigma_y) = \mathrm{det}(\Sigma_x)(\Sigma_y). $$

  1. From the block structure of $\tilde{\Sigma}$, we have

$$ \tilde{\Sigma}^{-1} = \begin{bmatrix} \Sigma_x^{-1} & \mathbf{0} \\ \mathbf{0} & \Sigma_y^{-1} \end{bmatrix} $$ which implies that $$ \Sigma \tilde{\Sigma}^{-1} = \begin{bmatrix} \Sigma_x & \Sigma' \\ \Sigma' & \Sigma_y \end{bmatrix} \begin{bmatrix} \Sigma_x^{-1} & \mathbf{0} \\ \mathbf{0} & \Sigma_y^{-1} \end{bmatrix} = \begin{bmatrix} \Sigma_x\Sigma_x^{-1} & \Sigma'\Sigma_y \\ \Sigma'\Sigma_x & \Sigma_y^{-1} \end{bmatrix} = \begin{bmatrix} \mathbf{1}_{n_1} & \Sigma'\Sigma_y \\ \Sigma'\Sigma_x & \mathbf{1}_{n_2} \end{bmatrix} $$

where $n_1$ and $n_2$ are positive integers such that $n_1 + n_2 = n$. Thus, it follows that $$ \mathrm{tr}\left( \Sigma \tilde{\Sigma}^{-1} \right) - n = n - n = 0. $$

  1. Since $\mu = \tilde{\mu}$, the right-most term in the expression for $D_{\rm KL}$ is equal to zero:

$$ \frac{1}{2}(\mu - \tilde{\mu})^\top\tilde{\Sigma}^{-1}(\mu - \tilde{\mu}) = 0. $$

Using these facts, we have shown that

$$ I(X:Y) = D_{\rm KL}((\mu,\Sigma) \mid\mid (\tilde{\mu}, \tilde{\Sigma})) = \frac{1}{2} \left[ \log\left( \frac{\mathrm{det}(\Sigma_x)\mathrm{det}(\Sigma_y)}{\mathrm{det}(\Sigma)} \right) \right] $$

as desired.


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