# Differential entropy

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Differential entropy

Differential entropy (also referred to as continuous entropy) is a concept in information theory that extends the idea of (Shannon) entropy, a measure of average surprisal of a random variable, to continuous probability distributions.

## Definition

Let X be a random variable with a probability density function f whose support is a set $\mathbb X$. The differential entropy h(X) or h(f) is defined as

$h(X) = -\int_\mathbb{X} f(x)\log f(x)\,dx$.

As with its discrete analog, the units of differential entropy depend on the base of the logarithm, which is usually 2 (i.e., the units are bits). See logarithmic units for logarithms taken in different bases. Related concepts such as joint, conditional differential entropy, and relative entropy are defined in a similar fashion.

One must take care in trying to apply properties of discrete entropy to differential entropy, since probability density functions can be greater than 1. For example, Uniform(0,1/2) has negative differential entropy $\int_0^\frac{1}{2} -2\log2\,dx=-\log2\,$.

Thus, differential entropy does not share all properties of discrete entropy.

Note that the continuous mutual information I(X;Y) has the distinction of retaining its fundamental significance as a measure of discrete information since it is actually the limit of the discrete mutual information of partitions of X and Y as these partitions become finer and finer. Thus it is invariant under non-linear homeomorphisms (continuous and uniquely invertible maps) [1], including linear [2] transformations of X and Y, and still represents the amount of discrete information that can be transmitted over a channel that admits a continuous space of values.

## Properties of differential entropy

• For two densities f and g, $D(f||g) \geq 0$ with equality if f = g almost everywhere. Similarly, for two random variables X and Y, $I(X;Y) \geq 0$ and $h(X|Y) \leq h(X)$ with equality if and only if X and Y are independent.
• The chain rule for differential entropy holds as in the discrete case
$h(X_1, \ldots, X_n) = \sum_{i=1}^{n} h(X_i|X_1, \ldots, X_{i-1}) \leq \sum h(X_i)$.
• Differential entropy is translation invariant, ie, h(X + c) = h(X) for a constant c.
• Differential entropy is in general not invariant under arbitrary invertible maps. In particular, for a constant a, $h(aX) = h(X) + \log \left| a \right|$. For a vector valued random variable X and a matrix A, $h(A\mathbf{X}) = h(\mathbf{X}) + \log \left| \det A \right|$.
• In general, for a transformation from a random vector X to a random vector with same dimension Y $\mathbf{Y} = m(\mathbf{X})$, the corresponding entropies are related via $h(\mathbf{Y}) \leq h(\mathbf{X}) + \int f(x) \log \left\vert \frac{\partial m}{\partial x} \right\vert dx$ where $\left\vert \frac{\partial m}{\partial x} \right\vert$ is the Jacobian of the transformation m. Equality is achieved if the transform is bijective, i.e., invertible.
• If a random vector $\mathbf{X} \in \mathbb{R}^{n}$ has mean zero and covariance matrix K, $h(\mathbf{X}) \leq \frac{1}{2} \log[(2\pi e)^n \det{K}]$ with equality if and only if X is jointly gaussian.

However, differential entropy does not have other desirable properties:

A modification of differential entropy that addresses this is the relative information entropy, also known as the Kullback–Leibler divergence, which includes an invariant measure factor (see limiting density of discrete points).

## Maximization in the normal distribution

With a normal distribution, differential entropy is maximized for a given variance. The following is a proof that a Gaussian variable has the largest entropy amongst all random variables of equal variance.

Let g(x) be a Gaussian PDF with mean μ and variance σ2 and f(x) an arbitrary PDF with the same variance. Since differential entropy is translation invariant we can assume that f(x) has the same mean of μ as g(x).

Consider the Kullback-Leibler divergence between the two distributions

\begin{align} 0 \leq D_{KL}(f || g) &= \int_{-\infty}^\infty f(x) \log \left( \frac{f(x)}{g(x)} \right) dx \\ &= -h_{f(x)} - \int_{-\infty}^\infty f(x)\log(g(x)) dx. \end{align} \!

Now note that

\begin{align} \int_{-\infty}^\infty f(x)\log(g(x)) dx &= \int_{-\infty}^\infty f(x)\log\left( \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\right) dx \\ &= \int_{-\infty}^\infty f(x) \log\frac{1}{\sqrt{2\pi\sigma^2}} dx + \int_{-\infty}^\infty f(x)\left( -\frac{(x-\mu)^2}{2\sigma^2}\right) dx \\ &= -\frac{1}{2}\log(2\pi\sigma^2) - \frac{\sigma^2}{2\sigma^2} = -\frac{1}{2}\left(\log(2\pi\sigma^2) + 1\right) \\ &= -h_{g(x)} \end{align} \!

because the result does not depend on f(x) other than through the variance. Combining the two results yields

$h_{g(x)} - h_{f(x)} \geq 0 \!$

with equality when g(x) = f(x) following from the properties of Kullback-Leibler divergence.

This result may also be demonstrated using the variational calculus. A Lagrangian function with two Lagrangian multipliers may be defined as:

$L=\int_{-\infty}^\infty g(x)\ln(g(x))\,dx+\lambda_0\left(1-\int_{-\infty}^\infty g(x)\,dx\right)+\lambda\left(\sigma^2-\int_{-\infty}^\infty g(x)(x-\mu)^2\,dx\right)$

where g(x) is some function with mean μ. When the entropy of g(x) is at a maximum and the constraint equations, which consist of the normalization condition $\left(1=\int_{-\infty}^\infty g(x)\,dx\right)$ and the requirement of fixed variance $\left(\sigma^2=\int_{-\infty}^\infty g(x)(x-\mu)^2\,dx\right)$, are both satisfied, then a small variation δg(x) about g(x) will produce a variation δL about L which is equal to zero:

$0=\delta L=\int_{-\infty}^\infty \delta g(x)\left[\ln(g(x))+1+\lambda_0+\lambda(x-\mu)^2\right]\,dx$

Since this must hold for any small δg(x), the term in brackets must be zero, and solving for g(x) yields:

$g(x)=e^{-\lambda_0-1-\lambda(x-\mu)^2}$

Using the constraint equations to solve for λ0 and λ yields the normal distribution:

$g(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$

## Example: Exponential distribution

Let X be an exponentially distributed random variable with parameter λ, that is, with probability density function

$f(x) = \lambda e^{-\lambda x} \mbox{ for } x \geq 0.$

Its differential entropy is then

 $h_e(X)\,$ $=-\int_0^\infty \lambda e^{-\lambda x} \log (\lambda e^{-\lambda x})\,dx$ $= -\left(\int_0^\infty (\log \lambda)\lambda e^{-\lambda x}\,dx + \int_0^\infty (-\lambda x) \lambda e^{-\lambda x}\,dx\right)$ $= -\log \lambda \int_0^\infty f(x)\,dx + \lambda E[X]$ $= -\log\lambda + 1\,.$

Here, he(X) was used rather than h(X) to make it explicit that the logarithm was taken to base e, to simplify the calculation.

## Differential entropies for various distributions

In the table below, $\Gamma(x) = \int_0^{\infty} e^{-t} t^{x-1} dt$ (the gamma function), $\psi(x) = \frac{d}{dx} \Gamma(x)$, $B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$, and γE is Euler's constant. Each distribution maximizes the entropy for a particular set of functional constraints listed in the fourth column, and the constraint that x be included in the support of the probability density, which is listed in the fifth column[3].

Table of differential entropies and corresponding maximum entropy constraints
Distribution Name Probability density function (pdf) Entropy in nats Maximum Entropy Constraint Support
Uniform $f(x) = \frac{1}{b-a}$ $\ln(b - a) \,$ None $[a,b]\,$
Normal $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$ $\ln\left(\sigma\sqrt{2\,\pi\,e}\right)$ $E(x)=\mu,\,E((x-\mu)^2)=\sigma^2$ $(-\infty,\infty)\,$
Exponential $f(x) = \lambda \exp\left(-\lambda x\right)$ $1 - \ln \lambda \,$ $E(x)=1/\lambda\,$ $[0,\infty)\,$
Rayleigh $f(x) = \frac{x}{\sigma^2} \exp\left(-\frac{x^2}{2\sigma^2}\right)$ $1 + \ln \frac{\sigma}{\sqrt{2}} + \frac{\gamma_E}{2}$ $E(x^2)=2\sigma^2, E(\ln(x))=\frac{\ln(2\sigma^2)-\gamma_E}{2}\,$ $[0,\infty)\,$
Beta $f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}$ for $0 \leq x \leq 1$ $\ln B(\alpha,\beta) - (\alpha-1)[\psi(\alpha) - \psi(\alpha +\beta)]\,$
$- (\beta-1)[\psi(\beta) - \psi(\alpha + \beta)] \,$
$E(\ln(x))=\psi(\alpha)-\psi(\alpha+\beta)\,$
$E(\ln(1-x))=\psi(\beta )-\psi(\alpha+\beta)\,$
$[0,1]\,$
Cauchy $f(x) = \frac{\gamma}{\pi} \frac{1}{\gamma^2 + x^2}$ $\ln(4\pi\gamma) \,$ $E(\ln(x^2+\gamma^2))=\ln(4\gamma^2)\,$ $(-\infty,\infty)\,$
Chi $f(x) = \frac{2}{2^{k/2} \Gamma(k/2)} x^{k-1} \exp\left(-\frac{x^2}{2}\right)$ $\ln{\frac{\Gamma(k/2)}{\sqrt{2}}} - \frac{k-1}{2} \psi\left(\frac{k}{2}\right) + \frac{k}{2}$ $E(x^2)=k,\,E(\ln(x))=\frac{1}{2}\left[\psi\left(\frac{k}{2}\right)\!+\!\ln(2)\right]$ $[0,\infty)\,$
Chi-squared $f(x) = \frac{1}{2^{k/2} \Gamma(k/2)} x^{\frac{k}{2}\!-\!1} \exp\left(-\frac{x}{2}\right)$ $\ln 2\Gamma\left(\frac{k}{2}\right) - \left(1 - \frac{k}{2}\right)\psi\left(\frac{k}{2}\right) + \frac{k}{2}$ $E(x)=k,\,E(\ln(x))=\psi\left(\frac{k}{2}\right)+\ln(2)$ $[0,\infty)\,$
Erlang $f(x) = \frac{\lambda^k}{(k-1)!} x^{k-1} \exp(-\lambda x)$ $(1-k)\psi(k) + \ln \frac{\Gamma(k)}{\lambda} + k$ $E(x)=k/\lambda,\,E(\ln(x))=\psi(k)-\ln(\lambda)$ $[0,\infty)\,$
F $f(x) = \frac{n_1^{\frac{n_1}{2}} n_2^{\frac{n_2}{2}}}{B(\frac{n_1}{2},\frac{n_2}{2})} \frac{x^{\frac{n_1}{2} - 1}}{(n_2 + n_1 x)^{\frac{n_1 + n2}{2}}}$ $\ln \frac{n_1}{n_2} B\left(\frac{n_1}{2},\frac{n_2}{2}\right) + \left(1 - \frac{n_1}{2}\right) \psi\left(\frac{n_1}{2}\right) -$
$\left(1 + \frac{n_2}{2}\right)\psi\left(\frac{n_2}{2}\right) + \frac{n_1 + n_2}{2} \psi\left(\frac{n_1\!+\!n_2}{2}\right)$
$\,$ $[0,\infty)\,$
Gamma $f(x) = \frac{x^{k - 1} \exp(-\frac{x}{\theta})}{\theta^k \Gamma(k)}$ $\ln(\theta \Gamma(k)) + (1 - k)\psi(k) + k \,$ $E(x)=k\theta,\,E(\ln(x))=\psi(k)+\ln(\theta)$ $[0,\infty)\,$
Laplace $f(x) = \frac{1}{2b} \exp\left(-\frac{|x - \mu|}{b}\right)$ $1 + \ln(2b) \,$ $E(|x-\mu|)=b\,$ $(-\infty,\infty)\,$
Logistic $f(x) = \frac{e^{-x}}{(1 + e^{-x})^2}$ $2 \,$ $\,$ $(-\infty,\infty)\,$
Lognormal $f(x) = \frac{1}{\sigma x \sqrt{2\pi}} \exp\left(-\frac{(\ln x - \mu)^2}{2\sigma^2}\right)$ $\mu + \frac{1}{2} \ln(2\pi e \sigma^2)$ $E(\ln(x))=\mu,E((\ln(x) - \mu)^2)=\sigma^2\,$ $[0,\infty)\,$
Maxwell-Boltzmann $f(x) = \frac{1}{a^3}\sqrt{\frac{2}{\pi}}\,x^{2}\exp\left(-\frac{x^2}{2a^2}\right)$ $\frac{1}{2}-\gamma_E-\ln(a\sqrt{2\pi})$ $E(x^2)=3a^2,\,E(\ln(x))\!=\!1\!+\!\ln\left(\frac{a}{\sqrt{2}}\right)\!-\!\frac{\gamma_E}{2}$ $[0,\infty)\,$
Generalized normal $f(x) = \frac{2 \beta^{\frac{\alpha}{2}}}{\Gamma(\frac{\alpha}{2})} x^{\alpha - 1} \exp(-\beta x^2)$ $\ln{\frac{\Gamma(\alpha/2)}{2\beta^{\frac{1}{2}}}} - \frac{\alpha - 1}{2} \psi\left(\frac{\alpha}{2}\right) + \frac{\alpha}{2}$ $\,$ $(-\infty,\infty)\,$
Pareto $f(x) = \frac{\alpha x_m^\alpha}{x^{\alpha+1}}$ $\ln \frac{x_m}{\alpha} + 1 + \frac{1}{\alpha}$ $E(\ln(x))=\frac{1}{\alpha}+\ln(x_m)\,$ $[x_m,\infty)\,$
Student's t $f(x) = \frac{(1 + x^2/\nu)^{-\frac{\nu+1}{2}}}{\sqrt{\nu}B(\frac{1}{2},\frac{\nu}{2})}$ $\frac{\nu\!+\!1}{2}\psi\left(\frac{\nu\!+\!1}{2}\right)\!-\!\psi\left(\frac{\nu}{2}\right)\!+\!\ln \sqrt{\nu} B\left(\frac{1}{2},\frac{\nu}{2}\right)$ $E(\ln(x^2\!+\!\nu))=\log \left(\nu\right)\!-\!\psi \left(\frac{\nu}{2}\right)\!+\!\psi\left(\frac{\nu\!+\!1}{2} \right)\,$ $(-\infty,\infty)\,$
Triangular $f(x) = \begin{cases} \frac{2x}{a} & 0 \leq x \leq a\\ \frac{2(1-x)}{1-a} & a \leq x \leq 1 \end{cases}$ $\frac{1}{2} - \ln 2$ $\,$ $[0,1]\,$
Weibull $f(x) = \frac{k}{\lambda^k} x^{k-1} \exp\left(-\frac{x^k}{\lambda^k}\right)$ $\frac{(k-1)\gamma_E}{k} + \ln \frac{\lambda}{k} + 1$ $E(x^k)=\lambda^k,E(\ln(x))=\ln(\lambda)-\frac{\gamma_E}{k}\,$ $[0,\infty)\,$
Multivariate normal $f_X(\vec{x}) =$
$\frac{\exp \left( -\frac{1}{2} ( \vec{x} - \vec{\mu})^\top \Sigma^{-1}\cdot(\vec{x} - \vec{\mu}) \right)} {(2\pi)^{N/2} \left|\Sigma\right|^{1/2}}$
$\frac{1}{2}\ln\{(2\pi e)^{N} \det(\Sigma)\}$ $\,$ $(-\vec{\infty},\vec{\infty})\,$

(Many of the differential entropies are from [4].

## Variants

As described above, differential entropy does not share all properties of discrete entropy. A modification of differential entropy adds an invariant measure factor to correct this, (see limiting density of discrete points). If m(x) is further constrained to be a probability density, the resulting notion is called relative entropy in information theory:

$D(p||m) = \int p(x)\log\frac{p(x)}{m(x)}\,dx.$

The definition of differential entropy above can be obtained by partitioning the range of X into bins of length h with associated sample points ih within the bins, for X Riemann integrable. This gives a quantized version of X, defined by Xh = ih if $ih \leq X \leq (i+1)h$. Then the entropy of Xh is

$H_h=-\sum_i hf(ih)\log f(ih) - \sum hf(ih)\log h$.

The first term on the right approximates the differential entropy, while the second term is approximately − log(h). Note that this procedure suggests that the entropy in the discrete sense of a continuous random variable should be $\infty$.

## References

1. ^ Kraskov, Alexander; Stögbauer, Grassberger (2004). "Estimating mutual information". Phys. Rev. E 60: 066138. arXiv:cond-mat/0305641. Bibcode 2004PhRvE..69f6138K. doi:10.1103/PhysRevE.69.066138.
2. ^ Fazlollah M. Reza (1961, 1994). An Introduction to Information Theory. Dover Publications, Inc., New York. ISBN 0-486-68210-2.
3. ^ Park, Sung Y.; Bera, Anil K. (2009). "Maximum entropy autoregressive conditional heteroskedasticity model". Journal of Econometrics (Elsevier): 219–230. Retrieved 2011-06-02.
4. ^ Lazo, A. and P. Rathie. On the entropy of continuous probability distributions Information Theory, IEEE Transactions on, 1978. 24(1): p. 120-122.
• Thomas M. Cover, Joy A. Thomas. Elements of Information Theory New York: Wiley, 1991. ISBN 0-471-06259-6

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