1
$\begingroup$

I would like to discuss the computation of the Tsallis entropy for the generalized Gaussian distribution. From the paper in the link https://www.sciencedirect.com/science/article/pii/S0167947322000822. The density function of the multivariate exponential power distribution, denoted as $MEP_m(s,\mu,\Sigma)$ , is defined on $\mathbb{R}^m$ and given by:

\begin{equation} f(x,;m,s,\mu,\Sigma) = \frac{\Gamma(m/2+1)}{\pi^{m/2}\Gamma(m/s+1)2^{m/s}\sqrt{\det \Sigma}} \exp\left(-\frac{1}{2}\Big[(x-\mu)^{ T}\Sigma^{-1}(x-\mu)\Big]^{s/2}\right), \end{equation}

In this equation, $\mu\in\mathbb{R}^m$ represents the mean vector, $\Sigma$ is a positive definite $m\times m$ matrix, $s>0$ is a shape parameter , and the variance-covariance matrix is denoted as $V =\beta\Sigma$, where the scale factor $\beta(m,s)$ is defined as:

\begin{equation} \beta(m,s) = \frac{2^{2/s}\Gamma\big[(m+2)/s\big]}{m\Gamma(m/s)}. \end{equation}

Additionally, we have the formula for Tsallis entropy, denoted as $H_q(f)$, as follows:

\begin{equation} H_q(f) = \frac{1}{1-q}\left(\int_{\mathbb{R}^{m}}f^{q}(x),dx-1\right), \quad q\neq 1. \end{equation}

My aim is to determine whether the Tsallis entropy is maximized by the generalized Gaussian distribution. How can we effectively compute and analyze the Tsallis entropy of this distribution in order to investigate its maximization properties? Any insights or methods to simplify this computation would be greatly appreciated. Thank you in advance for your help!

$\endgroup$
2
  • 2
    $\begingroup$ Since $f^q(x,m,s,\mu,\Sigma)$ is directly proportional to $f(x;m,qs,\mu,\Sigma)$ (with constant of proportionality independent of $x$), you don't have to calculate any more integrals: simply apply the formula you already have. $\endgroup$
    – whuber
    Commented Aug 24, 2023 at 16:20
  • $\begingroup$ Thank you so much for your help @whuber $\endgroup$
    – M.cadirci
    Commented Aug 24, 2023 at 20:05

1 Answer 1

2
$\begingroup$

Write

$$f(x;m,s,\mu,\Sigma) = \frac{1}{C(m,s,\Sigma)} \exp\left(-\frac{1}{2} h(x,\mu,\Sigma)^s\right).$$

The denominator is whatever quantity will make $f$ integrate to unity, whence

$$C(m,s,\Sigma) = \int_{\mathbb R^m} \exp\left(-\frac{1}{2} h(x,\mu,\Sigma)^{s}\right)\,\mathrm dx.$$

This is the only formula you need.

The integral involved in the entropy formula is

$$\begin{aligned} \int_{\mathbb R^m} f^q(x;m,s,\mu,\Sigma)\,\mathrm d x &= \frac{1}{C(m,s,\Sigma)^q} \int_{\mathbb R^m}\exp\left(-\frac{1}{2} h(x,\mu,\Sigma)^{sq}\right)\,\mathrm dx\\ &= \frac{C(m,sq,\Sigma)}{C(m,s,\Sigma)^q}. \end{aligned}$$

The rest is easy algebra.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.