Index: The Book of Statistical ProofsProbability Distributions ▷ Multivariate continuous distributions ▷ Normal-gamma distribution ▷ Differential entropy

Theorem: Let $x$ be an $n \times 1$ random vector and let $y$ be a positive random variable. Assume that $x$ and $y$ are jointly normal-gamma distributed:

$\label{eq:NG} (x,y) \sim \mathrm{NG}(\mu, \Lambda^{-1}, a, b)$

Then, the differential entropy of $x$ in nats is

$\label{eq:NG-dent} \begin{split} \mathrm{h}(x,y) &= \frac{n}{2} \ln(2\pi) - \frac{1}{2} \ln|\Lambda| + \frac{1}{2} n \\ &+ a + \ln \Gamma(a) - \frac{n-2+2a}{2} \psi(a) + \frac{n-2}{2} \ln b \; . \end{split}$ $\label{eq:NG-pdf} p(x,y) = p(x|y) \cdot p(y) = \mathcal{N}(x; \mu, (y \Lambda)^{-1}) \cdot \mathrm{Gam}(y; a, b) \; .$ $\label{eq:mvn-dent} \mathrm{h}(x) = \frac{n}{2} \ln(2\pi) + \frac{1}{2} \ln|\Sigma| + \frac{1}{2} n$ $\label{eq:gam-dent} \mathrm{h}(y) = a + \ln \Gamma(a) + (1-a) \cdot \psi(a) - \ln b$

where $\Gamma(x)$ is the gamma function and $\psi(x)$ is the digamma function.

The differential entropy of a continuous random variable in nats is given by

$\label{eq:dent} \mathrm{h}(Z) = - \int_{\mathcal{Z}} p(z) \ln p(z) \, \mathrm{d}z$

which, applied to the normal-gamma distribution over $x$ and $y$, yields

$\label{eq:NG-dent0} \mathrm{h}(x,y) = - \int_{0}^{\infty} \int_{\mathbb{R}^n} p(x,y) \, \ln p(x,y) \, \mathrm{d}x \, \mathrm{d}y \; .$

Using the law of conditional probability, this can be evaluated as follows:

$\label{eq:NG-dent1} \begin{split} \mathrm{h}(x,y) &= - \int_{0}^{\infty} \int_{\mathbb{R}^n} p(x|y) \, p(y) \, \ln p(x|y) \, p(y) \, \mathrm{d}x \, \mathrm{d}y \\ &= - \int_{0}^{\infty} \int_{\mathbb{R}^n} p(x|y)\, p(y) \, \ln p(x|y) \, \mathrm{d}x \, \mathrm{d}y - \int_{0}^{\infty} \int_{\mathbb{R}^n} p(x|y)\, p(y) \, \ln p(y) \, \mathrm{d}x \, \mathrm{d}y \\ &= \int_{0}^{\infty} p(y) \int_{\mathbb{R}^n} p(x|y) \, \ln p(x|y) \, \mathrm{d}x \, \mathrm{d}y + \int_{0}^{\infty} p(y) \, \ln p(y) \int_{\mathbb{R}^n} p(x|y) \, \mathrm{d}x \, \mathrm{d}y \\ &= \left\langle \mathrm{h}(x|y) \right\rangle_{p(y)} + \mathrm{h}(y) \; . \end{split}$

In other words, the differential entropy of the normal-gamma distribution over $x$ and $y$ is equal to the sum of a multivariate normal entropy regarding $x$ conditional on $y$, expected over $y$, and a univariate gamma entropy regarding $y$.

From equations \eqref{eq:NG-pdf} and \eqref{eq:mvn-dent}, the first term becomes

$\label{eq:exp-mvn-dent-s1} \begin{split} \left\langle \mathrm{h}(x|y) \right\rangle_{p(y)} &= \left\langle \frac{n}{2} \ln(2\pi) + \frac{1}{2} \ln|(y \Lambda)^{-1}| + \frac{1}{2} n \right\rangle_{p(y)} \\ &= \left\langle \frac{n}{2} \ln(2\pi) - \frac{1}{2} \ln|(y \Lambda)| + \frac{1}{2} n \right\rangle_{p(y)} \\ &= \left\langle \frac{n}{2} \ln(2\pi) - \frac{1}{2} \ln(y^n |\Lambda|) + \frac{1}{2} n \right\rangle_{p(y)} \\ &= \frac{n}{2} \ln(2\pi) - \frac{1}{2} \ln|\Lambda| + \frac{1}{2} n - \left\langle \frac{n}{2} \ln y \right\rangle_{p(y)} \\ \end{split}$

and using the relation $y \sim \mathrm{Gam}(a,b) \Rightarrow \left\langle \ln y \right\rangle = \psi(a) - \ln(b)$, we have

$\label{eq:exp-mvn-dent-s2} \begin{split} \left\langle \mathrm{h}(x|y) \right\rangle_{p(y)} = \frac{n}{2} \ln(2\pi) - \frac{1}{2} \ln|\Lambda| + \frac{1}{2} n - \frac{n}{2} \psi(a) + \frac{n}{2} \ln b \; . \end{split}$

By plugging \eqref{eq:exp-mvn-dent-s2} and \eqref{eq:gam-dent} into \eqref{eq:NG-dent1}, one arrives at the differential entropy given by \eqref{eq:NG-dent}.

Sources:

Metadata: ID: P238 | shortcut: ng-dent | author: JoramSoch | date: 2021-07-08, 10:51.