微分熵

此词条暂由Henry翻译。

Differential entropy (also referred to as continuous entropy) is a concept in information theory that began as an attempt by Shannon to extend the idea of (Shannon) entropy, a measure of average surprisal of a random variable, to continuous probability distributions. Unfortunately, Shannon did not derive this formula, and rather just assumed it was the correct continuous analogue of discrete entropy, but it is not. The actual continuous version of discrete entropy is the limiting density of discrete points (LDDP). Differential entropy (described here) is commonly encountered in the literature, but it is a limiting case of the LDDP, and one that loses its fundamental association with discrete entropy.

Differential entropy (also referred to as continuous entropy) is a concept in information theory that began as an attempt by Shannon to extend the idea of (Shannon) entropy, a measure of average surprisal of a random variable, to continuous probability distributions. Unfortunately, Shannon did not derive this formula, and rather just assumed it was the correct continuous analogue of discrete entropy, but it is not.

微分熵(也称为连续熵)是信息论中的一个概念，最初由香农尝试将(香农)熵的概念扩展到连续的概率分布，香农熵是衡量一个随机变量的平均惊人程度的指标。不幸的是，香农没有推导出这个公式，而只是假设它是离散熵的正确连续模拟，但事实上它不是。

[math]\displaystyle{ h(X_1, \ldots, X_n) = \sum_{i=1}^{n} h(X_i|X_1, \ldots, X_{i-1}) \leq \sum_{i=1}^{n} h(X_i) }[/math].

< math > h (x _ 1，ldots，xn) = sum _ { i = 1} ^ { n } h (x _ i | x _ 1，ldots，x _ { i-1}) leq sum _ { i = 1} ^ { n } h (x _ i) </math > .

Definition

定义 Let [math]\displaystyle{ X }[/math] be a random variable with a probability density function [math]\displaystyle{ f }[/math] whose support is a set [math]\displaystyle{ \mathcal X }[/math]. The differential entropy [math]\displaystyle{ h(X) }[/math] or [math]\displaystyle{ h(f) }[/math] is defined as

[math]\displaystyle{ h(X+c) = h(X) }[/math]

[ math > h (x + c) = h (x) </math >

{{Equation box 1

In particular, for a constant [math]\displaystyle{ a }[/math]

特别是对于一个常量

|indent =

[math]\displaystyle{ h(aX) = h(X)+ \log |a| }[/math]

H (aX) = h (x) + log | a | </math

|title=

For a vector valued random variable [math]\displaystyle{ \mathbf{X} }[/math] and an invertible (square) matrix [math]\displaystyle{ \mathbf{A} }[/math]

对于向量值随机变量 < math > mathbf { x } </math > 和可逆矩阵 < math > mathbf { a } </math >

|equation = [math]\displaystyle{ h(X) = -\int_\mathcal{X} f(x)\log f(x)\,dx }[/math]

[math]\displaystyle{ h(\mathbf{A}\mathbf{X})=h(\mathbf{X})+\log \left( |\det \mathbf{A}| \right) }[/math]

< math > h (mathbf { a } mathbf { x }) = h (mathbf { x }) + log left (| det mathbf { a } | right) </math >

|cellpadding= 6

|border

[math]\displaystyle{ h(\mathbf{Y}) \leq h(\mathbf{X}) + \int f(x) \log \left\vert \frac{\partial m}{\partial x} \right\vert dx }[/math]

[ math > h (mathbf { y }) leq h (mathbf { x }) + int f (x) log left vert frac { partial m }{ partial x } right vert dx </math >

|border colour = #0073CF

where [math]\displaystyle{ \left\vert \frac{\partial m}{\partial x} \right\vert }[/math] is the Jacobian of the transformation [math]\displaystyle{ m }[/math].

其中“ math” > “ left vert”{ partial m }{ partial x }“ right vert” >/math > 是变换的雅可比矩阵。

|background colour=#F5FFFA}}

However, differential entropy does not have other desirable properties:

然而，微分熵并没有其他令人满意的特性:

For probability distributions which don't have an explicit density function expression, but have an explicit quantile function expression, [math]\displaystyle{ Q(p) }[/math], then [math]\displaystyle{ h(Q) }[/math] can be defined in terms of the derivative of [math]\displaystyle{ Q(p) }[/math] i.e. the quantile density function [math]\displaystyle{ Q'(p) }[/math] as

[math]\displaystyle{ h(Q) = \int_0^1 \log Q'(p)\,dp }[/math].

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

针对这些缺点，微分熵的一个改进是相对熵，也被称为 Kullback-Leibler 分歧，其中包括一个不变测度因子(见离散点的极限密度)。

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. Unlike the discrete analog, the differential entropy has an offset that depends on the units used to measure [math]\displaystyle{ X }[/math]. For example, the differential entropy of a quantity measured in millimeters will be 模板:Not a typo more than the same quantity measured in meters; a dimensionless quantity will have differential entropy of 模板:Not a typo more than the same quantity divided by 1000.

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, the uniform distribution [math]\displaystyle{ \mathcal{U}(0,1/2) }[/math] has negative differential entropy

With a normal distribution, differential entropy is maximized for a given variance. A Gaussian random variable has the largest entropy amongst all random variables of equal variance, or, alternatively, the maximum entropy distribution under constraints of mean and variance is the Gaussian.

在一个正态分布下，对于给定的方差，微分熵是最大的。在所有方差相等的随机变量中，高斯型随机变量的熵最大，或者说在均值和方差约束下的最大熵分布是高斯型随机变量。

[math]\displaystyle{ \int_0^\frac{1}{2} -2\log(2)\,dx=-\log(2)\, }[/math].

Let [math]\displaystyle{ g(x) }[/math] be a Gaussian PDF with mean μ and variance [math]\displaystyle{ \sigma^2 }[/math] and [math]\displaystyle{ f(x) }[/math] an arbitrary PDF with the same variance. Since differential entropy is translation invariant we can assume that [math]\displaystyle{ f(x) }[/math] has the same mean of [math]\displaystyle{ \mu }[/math] as [math]\displaystyle{ g(x) }[/math].

设g(x) 是一个高斯分布的 PDF，平均值μ 和方差σ2和f(x)一个任意的 PDF，方差相同。由于微分熵是平移不变的，我们可以假设 f(x) 与g(x）具有相同的平均值。

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

Consider the Kullback–Leibler divergence between the two distributions

考虑两个分布之间的 Kullback-Leibler 散度

Note that the continuous mutual information [math]\displaystyle{ I(X;Y) }[/math] 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 [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] as these partitions become finer and finer. Thus it is invariant under non-linear homeomorphisms (continuous and uniquely invertible maps), including linear transformations of [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math], and still represents the amount of discrete information that can be transmitted over a channel that admits a continuous space of values.

because the result does not depend on [math]\displaystyle{ f(x) }[/math] other than through the variance. Combining the two results yields

因为结果并不依赖于f(x)，而是通过方差。将这两个结果结合起来就会产生结果

[math]\displaystyle{ h(g) - h(f) \geq 0 \! }[/math]

[数学]-[数学]

For the direct analogue of discrete entropy extended to the continuous space, see limiting density of discrete points.

with equality when [math]\displaystyle{ f(x)=g(x) }[/math] following from the properties of Kullback–Leibler divergence.

当f (x) = g (x)遵循 Kullback-Leibler 分歧的性质时。

Properties of differential entropy

微分熵的性质

• For probability densities [math]\displaystyle{ f }[/math] and [math]\displaystyle{ g }[/math], the Kullback–Leibler divergence [math]\displaystyle{ D_{KL}(f || g) }[/math] is greater than or equal to 0 with equality only if [math]\displaystyle{ f=g }[/math] almost everywhere. Similarly, for two random variables [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math], [math]\displaystyle{ I(X;Y) \ge 0 }[/math] and [math]\displaystyle{ h(X|Y) \le h(X) }[/math] with equality if and only if [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] are independent.

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

这个结果也可以用变分法来证明。具有两个拉格朗日乘数的拉格朗日函数可定义为:

• The chain rule for differential entropy holds as in the discrete case

[math]\displaystyle{ h(X_1, \ldots, X_n) = \sum_{i=1}^{n} h(X_i|X_1, \ldots, X_{i-1}) \leq \sum_{i=1}^{n} h(X_i) }[/math].

[math]\displaystyle{ 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) }[/math]

< math > l = int _ {-infty } ^ infty g (x) ln (g (x)) ，dx-lambda _ 0 left (1-int _ {-infty } ^ infty g (x) ，dx 右)-lambda left (sigma ^ 2-int _ {-infty } ^ infty g (x)(x-mu) ^ 2，dx 右) </math >

• Differential entropy is translation invariant, i.e. for a constant [math]\displaystyle{ c }[/math].

[math]\displaystyle{ h(X+c) = h(X) }[/math]

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 [math]\displaystyle{ \left(1=\int_{-\infty}^\infty g(x)\,dx\right) }[/math] and the requirement of fixed variance [math]\displaystyle{ \left(\sigma^2=\int_{-\infty}^\infty g(x)(x-\mu)^2\,dx\right) }[/math], are both satisfied, then a small variation δg(x) about g(x) will produce a variation δL about L which is equal to zero:

其中 g (x)是平均 μ 的函数。当 g (x)的熵处于最大值时，由归一化条件 1=∫∞−∞g(x)dx和固定方差σ2=∫∞−∞g(x)(x−μ)2dx组成的约束方程都满足时，那么关于 g (x)的一个小变化 δg (x)将产生一个等于零的关于L的变化δL:

• Differential entropy is in general not invariant under arbitrary invertible maps.

In particular, for a constant [math]\displaystyle{ a }[/math]

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

0 = delta l = int _ {-infty } ^ infty delta g (x) left (ln (g (x)) + 1 + lambda _ 0 + lambda (x-mu) ^ 2 right) ，dx </math >

[math]\displaystyle{ h(aX) = h(X)+ \log |a| }[/math]

For a vector valued random variable [math]\displaystyle{ \mathbf{X} }[/math] and an invertible (square) matrix [math]\displaystyle{ \mathbf{A} }[/math]

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

因为这对任何小的 δg (x)都成立，括号中的项必须为零，求 g (x)的结果是:

[math]\displaystyle{ h(\mathbf{A}\mathbf{X})=h(\mathbf{X})+\log \left( |\det \mathbf{A}| \right) }[/math]

• In general, for a transformation from a random vector to another random vector with same dimension [math]\displaystyle{ \mathbf{Y}=m \left(\mathbf{X}\right) }[/math], the corresponding entropies are related via

[math]\displaystyle{ g(x)=e^{-\lambda_0-1-\lambda(x-\mu)^2} }[/math]

< math > g (x) = e ^ {-lambda _ 0-1-lambda (x-mu) ^ 2} </math >

[math]\displaystyle{ h(\mathbf{Y}) \leq h(\mathbf{X}) + \int f(x) \log \left\vert \frac{\partial m}{\partial x} \right\vert dx }[/math]

where [math]\displaystyle{ \left\vert \frac{\partial m}{\partial x} \right\vert }[/math] is the Jacobian of the transformation [math]\displaystyle{ m }[/math]. The above inequality becomes an equality if the transform is a bijection. Furthermore, when [math]\displaystyle{ m }[/math] is a rigid rotation, translation, or combination thereof, the Jacobian determinant is always 1, and [math]\displaystyle{ h(Y)=h(X) }[/math].

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

用约束方程求解 λ < sub > 0 和 λ 得到正态分布:

• If a random vector [math]\displaystyle{ X \in \mathbb{R}^n }[/math] has mean zero and covariance matrix [math]\displaystyle{ K }[/math], [math]\displaystyle{ h(\mathbf{X}) \leq \frac{1}{2} \log(\det{2 \pi e K}) = \frac{1}{2} \log[(2\pi e)^n \det{K}] }[/math] with equality if and only if [math]\displaystyle{ X }[/math] is jointly gaussian (see below).

[math]\displaystyle{ g(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} }[/math]

< math > g (x) = frac {1}{ sqrt {2 pi sigma ^ 2} e ^ {-frac {(x-mu) ^ 2}{2 sigma ^ 2}} </math >

However, differential entropy does not have other desirable properties:

• It is not invariant under change of variables, and is therefore most useful with dimensionless variables.

• It can be negative.

Let [math]\displaystyle{ X }[/math] be an exponentially distributed random variable with parameter [math]\displaystyle{ \lambda }[/math], that is, with probability density function

设 x 是一个指数分布的随机变量，它的参数是 λ，也就是概率密度函数

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

[math]\displaystyle{ f(x) = \lambda e^{-\lambda x} \mbox{ for } x \geq 0. }[/math]

{ for } x geq 0. </math >

Maximization in the normal distribution

Theorem

Its differential entropy is then

它的微分熵就在那时

With a normal distribution, differential entropy is maximized for a given variance. A Gaussian random variable has the largest entropy amongst all random variables of equal variance, or, alternatively, the maximum entropy distribution under constraints of mean and variance is the Gaussian.

Proof

Consider the Kullback–Leibler divergence between the two distributions

[math]\displaystyle{ \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 \\ | \lt math\gt = -\log \lambda \int_0^\infty f(x)\,dx + \lambda E[X] }[/math]

&= -\tfrac{1}{2}\log(2\pi\sigma^2) - \log(e)\frac{\sigma^2}{2\sigma^2} \\

[math]\displaystyle{ h_e(X)\, }[/math]	< math > h _ e (x) ，</math > Let [math]\displaystyle{ g(x) }[/math] be a Gaussian PDF with mean μ and variance [math]\displaystyle{ \sigma^2 }[/math] and [math]\displaystyle{ f(x) }[/math] an arbitrary PDF with the same variance. Since differential entropy is translation invariant we can assume that [math]\displaystyle{ f(x) }[/math] has the same mean of [math]\displaystyle{ \mu }[/math] as [math]\displaystyle{ g(x) }[/math].	[math]\displaystyle{ =-\int_0^\infty \lambda e^{-\lambda x} \log (\lambda e^{-\lambda x})\,dx }[/math]	< math > =-int _ 0 ^ infty lambda e ^ {-lambda x } log (lambda e ^ {-lambda x }) ，dx </math >
	[math]\displaystyle{ 0 \leq D_{KL}(f || g) = \int_{-\infty}^\infty f(x) \log \left( \frac{f(x)}{g(x)} \right) dx = -h(f) - \int_{-\infty}^\infty f(x)\log(g(x)) dx. }[/math]	[math]\displaystyle{ = -\left(\int_0^\infty (\log \lambda)\lambda e^{-\lambda x}\,dx + \int_0^\infty (-\lambda x) \lambda e^{-\lambda x}\,dx\right) }[/math]	< math > =-left (int _ 0 ^ infty (log lambda) lambda e ^ {-lambda x } ，dx + int _ 0 ^ infty (- lambda x) lambda e ^ {-lambda x } ，dx right) </math > Now note that
< math > =-log lambda int _ 0 ^ infty f (x) ，dx + lambda e [ x ] </math > &= \int_{-\infty}^\infty f(x) \log\frac{1}{\sqrt{2\pi\sigma^2}} dx + \log(e)\int_{-\infty}^\infty f(x)\left( -\frac{(x-\mu)^2}{2\sigma^2}\right) dx \\
	&= -\tfrac{1}{2}\left(\log(2\pi\sigma^2) + \log(e)\right) \\	[math]\displaystyle{ = -\log\lambda + 1\,. }[/math]	< math > =-log lambda + 1，. </math > &= -\tfrac{1}{2}\log(2\pi e \sigma^2) \\

&= -h(g)

\end{align}</math>

Here, [math]\displaystyle{ h_e(X) }[/math] was used rather than [math]\displaystyle{ h(X) }[/math] to make it explicit that the logarithm was taken to base e, to simplify the calculation.

在这里，使用 < math > h _ e (x) </math > 而不是 < math > h (x) </math > 来明确对数是以 e 为底，以简化计算。

because the result does not depend on [math]\displaystyle{ f(x) }[/math] other than through the variance. Combining the two results yields

[math]\displaystyle{ h(g) - h(f) \geq 0 \! }[/math]

with equality when [math]\displaystyle{ f(x)=g(x) }[/math] following from the properties of Kullback–Leibler divergence.

The differential entropy yields a lower bound on the expected squared error of an estimator. For any random variable [math]\displaystyle{ X }[/math] and estimator [math]\displaystyle{ \widehat{X} }[/math] the following holds:

对于估计量的预期平方误差，微分熵产生一个下限。对于任何随机变量 < math > x </math > 和估计量 < math > widedhat { x } </math > 下面的值:

[math]\displaystyle{ \operatorname{E}[(X - \widehat{X})^2] \ge \frac{1}{2\pi e}e^{2h(X)} }[/math]

(x-widehat { x }) ^ 2] ge frac {1}{2 pi e } e ^ {2 h (x)} </math >

Alternative proof

with equality if and only if [math]\displaystyle{ X }[/math] is a Gaussian random variable and [math]\displaystyle{ \widehat{X} }[/math] is the mean of [math]\displaystyle{ X }[/math].

当且仅当 < math > x </math > 是一个 Gaussian 随机变量，而 < math > x } </math > 是 < math > x </math > 的平均值。

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

[math]\displaystyle{ 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) }[/math]

In the table below [math]\displaystyle{ \Gamma(x) = \int_0^{\infty} e^{-t} t^{x-1} dt }[/math] is the gamma function, [math]\displaystyle{ \psi(x) = \frac{d}{dx} \ln\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)} }[/math] is the digamma function, [math]\displaystyle{ B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} }[/math] is the beta function, and γ_E is Euler's constant.[math]\displaystyle{ - (\beta-1)[\psi(\beta) - \psi(\alpha + \beta)] \, }[/math]||[math]\displaystyle{ [0,1]\, }[/math]

在下面的表格中，Gamma (x) = int _ 0 ^ { infty } e ^ {-t } t ^ { x-1} dt </math > 是 Gamma 函数,{ math > psi (x) = frac { d }{ dx } ln Gamma (x) = frac { Gamma’(x)}{ Gamma (x)} </math > 是双伽玛函数,b (p，q) = frac { Gamma (p) Gamma (q)}{ Gamma (p + q)} </math > 是 β 函数，γ < sub > e 是欧拉常数。[ math ]-(beta-1)[ psi (beta)-psi (alpha + beta)] | | < math > [0,1] ，</math >

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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 [math]\displaystyle{ \left(1=\int_{-\infty}^\infty g(x)\,dx\right) }[/math] and the requirement of fixed variance [math]\displaystyle{ \left(\sigma^2=\int_{-\infty}^\infty g(x)(x-\mu)^2\,dx\right) }[/math], are both satisfied, then a small variation δg(x) about g(x) will produce a variation δL about L which is equal to zero:

| Cauchy || [math]\displaystyle{ f(x) = \frac{\gamma}{\pi} \frac{1}{\gamma^2 + x^2} }[/math] || [math]\displaystyle{ \ln(4\pi\gamma) \, }[/math]||[math]\displaystyle{ (-\infty,\infty)\, }[/math]

| Cauchy | | < math > f (x) = frac { gamma }{ pi }{ pi ^ 2 + x ^ 2} </math > | < math > ln (4pi gamma) ，</math > | < math > (- infty，infty) ，</math >

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

| Chi || [math]\displaystyle{ f(x) = \frac{2}{2^{k/2} \Gamma(k/2)} x^{k-1} \exp\left(-\frac{x^2}{2}\right) }[/math] || [math]\displaystyle{ \ln{\frac{\Gamma(k/2)}{\sqrt{2}}} - \frac{k-1}{2} \psi\left(\frac{k}{2}\right) + \frac{k}{2} }[/math]||[math]\displaystyle{ [0,\infty)\, }[/math]

| Chi | | < math > f (x) = frac {2}{2 ^ { k/2} Gamma (k/2)}} x ^ { k-1} exp left (- frac { x ^ 2}{2}右) </math > | < math > ln { frac {(k/2)}}}{2}}}-frac {2} psi (frac { k }{2}右) + frac {2} </math > | | math > [0，infty) ，</math >

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

| Chi-squared || [math]\displaystyle{ f(x) = \frac{1}{2^{k/2} \Gamma(k/2)} x^{\frac{k}{2}\!-\!1} \exp\left(-\frac{x}{2}\right) }[/math] || [math]\displaystyle{ \ln 2\Gamma\left(\frac{k}{2}\right) - \left(1 - \frac{k}{2}\right)\psi\left(\frac{k}{2}\right) + \frac{k}{2} }[/math]||[math]\displaystyle{ [0,\infty)\, }[/math]

| Chi-squared | < math > f (x) = frac {1}{2 ^ { k/2} Gamma (k/2)} x ^ { frac { k }{2} !-! 1} exp left (- frac { x }{2}右) </math > | < math > | < math > ln 2 Gamma left (frac { k }{2}右)-left (1-frac { k }{2}右)左(frac { k }2}右) + c { k {2}{ infmath | < < math > [0，fraty) ，</math >

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[math]\displaystyle{ g(x)=e^{-\lambda_0-1-\lambda(x-\mu)^2} }[/math]

| Erlang || [math]\displaystyle{ f(x) = \frac{\lambda^k}{(k-1)!} x^{k-1} \exp(-\lambda x) }[/math] || [math]\displaystyle{ (1-k)\psi(k) + \ln \frac{\Gamma(k)}{\lambda} + k }[/math]||[math]\displaystyle{ [0,\infty)\, }[/math]

| Erlang | | < math > f (x) = frac { lambda ^ k }{(k-1) ! }X ^ { k-1} exp (- lambda x) </math > | < math > (1-k) psi (k) + ln frac { Gamma (k)}{ lambda } + k </math > | < math > [0，infty ] ，</math >

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

| F || [math]\displaystyle{ 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}}} }[/math] || [math]\displaystyle{ \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) - }[/math]
[math]\displaystyle{ \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) }[/math]||[math]\displaystyle{ [0,\infty)\, }[/math]

我们会找到你的| | < math > f (x) = frac{ n _ 1 ^ { frac { n _ 1}{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 + n _ 2}{2}}{2}{2}} </} </math > | | | (frac { n _ 1}{ n _ 2} b left (frac { n _ 1}{2} ,2}{2}{2}{2}{2}{2}{1}{2}{2}{2}{2}{2}{2}{2}{2}{2}{2}{1}{2}{2}{2}{2}{2}{2}{2}{2}{2}{2}{2}{2}{2}{2}{2}{2}{2}{2} psi 左(frac { n _ 1！+\！[0，infty) ，</math > | < math >

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[math]\displaystyle{ g(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} }[/math]

| Gamma || [math]\displaystyle{ f(x) = \frac{x^{k - 1} \exp(-\frac{x}{\theta})}{\theta^k \Gamma(k)} }[/math] || [math]\displaystyle{ \ln(\theta \Gamma(k)) + (1 - k)\psi(k) + k \, }[/math]||[math]\displaystyle{ [0,\infty)\, }[/math]

| Gamma | | < math > f (x) = frac { x ^ { k-1} exp (- frac { x }{ theta })}{ theta ^ k Gamma (k)} </math > | < math > ln (theta Gamma (k)) + (1-k) psi (k) + k，</math > | < math > [0，infty) ，</math >

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Example: Exponential distribution

| Laplace || [math]\displaystyle{ f(x) = \frac{1}{2b} \exp\left(-\frac{|x - \mu|}{b}\right) }[/math] || [math]\displaystyle{ 1 + \ln(2b) \, }[/math]||[math]\displaystyle{ (-\infty,\infty)\, }[/math]

| Laplace | | < math > f (x) = frac {1}{2b } exp left (- frac { | x-mu | }{ b } right) </math > | < math > 1 + ln (2b) ，</math > | < math > (- infty，infty) ，</math >

Let [math]\displaystyle{ X }[/math] be an exponentially distributed random variable with parameter [math]\displaystyle{ \lambda }[/math], that is, with probability density function

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| Logistic || [math]\displaystyle{ f(x) = \frac{e^{-x}}{(1 + e^{-x})^2} }[/math] || [math]\displaystyle{ 2 \, }[/math]||[math]\displaystyle{ (-\infty,\infty)\, }[/math]

| Logistic | | < math > f (x) = frac { e ^ {-x }{(1 + e ^ {-x }) ^ 2} </math > | < math > 2，</math > | < math > (- infty，infty) ，</math >

[math]\displaystyle{ f(x) = \lambda e^{-\lambda x} \mbox{ for } x \geq 0. }[/math]

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| Lognormal || [math]\displaystyle{ f(x) = \frac{1}{\sigma x \sqrt{2\pi}} \exp\left(-\frac{(\ln x - \mu)^2}{2\sigma^2}\right) }[/math] || [math]\displaystyle{ \mu + \frac{1}{2} \ln(2\pi e \sigma^2) }[/math]||[math]\displaystyle{ [0,\infty)\, }[/math]

| Lognormal | < math > f (x) = frac {1}{ sigma x sqrt {2 pi } exp left (- frac {(ln x-mu) ^ 2}{2 sigma ^ 2} right) </math > | < math > mu + frac {1}{2} ln (2 pi e sigma ^ 2) </math > | < math > [0，infty) ，</math >

Its differential entropy is then

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\frac{2(b-x)}{(b-a)(b-c)} & \mathrm{for\ } c < x \le b, \\[4pt] Frac {2(b-x)}{(b-a)(b-c)} & mathrm { for } c < x le b，[4 pt ]

Maxwell–Boltzmann	[math]\displaystyle{ f(x) = \frac{1}{a^3}\sqrt{\frac{2}{\pi}}\,x^{2}\exp\left(-\frac{x^2}{2a^2}\right) }[/math]	[math]\displaystyle{ \ln(a\sqrt{2\pi})+\gamma_E-\frac{1}{2} }[/math]	[math]\displaystyle{ [0,\infty)\, }[/math]	| < math > f (x) = frac {1}{ a ^ 3}{ frac {2}{ pi } ，x ^ {2} exp left (- frac { x ^ 2}{2a ^ 2}右) </math > | < math > ln (a sqrt {2 pi }) + e-frac {1} </math > | | math < 0，infty) ，</math >
[math]\displaystyle{ h_e(X)\, }[/math]	Generalized normal	[math]\displaystyle{ f(x) = \frac{2 \beta^{\frac{\alpha}{2}}}{\Gamma(\frac{\alpha}{2})} x^{\alpha - 1} \exp(-\beta x^2) }[/math]	[math]\displaystyle{ \ln{\frac{\Gamma(\alpha/2)}{2\beta^{\frac{1}{2}}}} - \frac{\alpha - 1}{2} \psi\left(\frac{\alpha}{2}\right) + \frac{\alpha}{2} }[/math]	[math]\displaystyle{ (-\infty,\infty)\, }[/math]	| < math > f (x) = frac{2 beta ^ { frac { alpha }{2}{ Gamma (frac { alpha }{2})} x ^ { alpha-1} exp (- beta x ^ 2) </math > | | < math > ln { frac { Gamma (alpha/2)}{2 beta ^ { frac {1}{2}}}}}-frac { alpha-1}{2} psi left (frac { alpha }{2} right) + frac { alpha }{2}}{2}| | < math > (- infty，infty) ，</math >	[math]\displaystyle{ =-\int_0^\infty \lambda e^{-\lambda x} \log (\lambda e^{-\lambda x})\,dx }[/math]
Pareto	[math]\displaystyle{ f(x) = \frac{\alpha x_m^\alpha}{x^{\alpha+1}} }[/math]	[math]\displaystyle{ \ln \frac{x_m}{\alpha} + 1 + \frac{1}{\alpha} }[/math]	[math]\displaystyle{ [x_m,\infty)\, }[/math]	< math > f (x) = frac { alpha x _ m ^ alpha }{ x ^ { alpha + 1}} </math > | < math > ln frac { x _ m }{ alpha } + 1 + frac {1}{ alpha } </math > | < math > [ x _ m，infty ] ，</math >
[math]\displaystyle{ = -\left(\int_0^\infty (\log \lambda)\lambda e^{-\lambda x}\,dx + \int_0^\infty (-\lambda x) \lambda e^{-\lambda x}\,dx\right) }[/math]	Student's t	[math]\displaystyle{ f(x) = \frac{(1 + x^2/\nu)^{-\frac{\nu+1}{2}}}{\sqrt{\nu}B(\frac{1}{2},\frac{\nu}{2})} }[/math]	[math]\displaystyle{ \frac{\nu\!+\!1}{2}\left(\psi\left(\frac{\nu\!+\!1}{2}\right)\!-\!\psi\left(\frac{\nu}{2}\right)\right)\!+\!\ln \sqrt{\nu} B\left(\frac{1}{2},\frac{\nu}{2}\right) }[/math]	[math]\displaystyle{ (-\infty,\infty)\, }[/math]	< math > f (x) = frac {(1 + x ^ 2/nu) ^ {-frac { nu + 1}{2}}{{ sqrt { nu } b (frac {1}{2} ，frac { nu }{2})} </math | | | < math > frac { nu! + ! 1}{2}右) !-! 左(psi (frac { nu! + 1}{2}右) !-! 左(frac { nu }{2右) ! + ! { nu }{ n 左(frac {2}右)
	Triangular	[math]\displaystyle{ f(x) = \begin{cases} | 三角形 | | \lt math \gt f (x) = begin { cases } | \lt math\gt = -\log \lambda \int_0^\infty f(x)\,dx + \lambda E[X] }[/math] \frac{2(x-a)}{(b-a)(c-a)} & \mathrm{for\ } a \le x \leq c, \\[4pt] Frac {2(x-a)}{(b-a)(c-a)} & mathrm { for } a le x leq c，[4 pt ]
\end{cases}</math> || [math]\displaystyle{ \frac{1}{2} + \ln \frac{b-a}{2} }[/math]||[math]\displaystyle{ [0,1]\, }[/math] 结束{ cases } </math > | | < math > frac {1}{2} + ln frac { b-a }{2} </math > | < math > [0,1] ，</math >	[math]\displaystyle{ = -\log\lambda + 1\,. }[/math]

| Weibull || [math]\displaystyle{ f(x) = \frac{k}{\lambda^k} x^{k-1} \exp\left(-\frac{x^k}{\lambda^k}\right) }[/math] || [math]\displaystyle{ \frac{(k-1)\gamma_E}{k} + \ln \frac{\lambda}{k} + 1 }[/math]||[math]\displaystyle{ [0,\infty)\, }[/math]

| Weibull | | < math > f (x) = frac { k }{ lambda ^ k } x ^ { k-1} exp left (- frac { x ^ k }{ lambda ^ k } right) </math > | < math > | < math > frac {(k-1) gamma _ e }{ k } + ln frac { lambda }{ k } + 1 </math > | < math > [0，infty) ，</math >

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Here, [math]\displaystyle{ h_e(X) }[/math] was used rather than [math]\displaystyle{ h(X) }[/math] to make it explicit that the logarithm was taken to base e, to simplify the calculation.

| Multivariate normal || [math]\displaystyle{ 多元正态 | | \lt 数学 \gt f_X(\vec{x}) = }[/math]
[math]\displaystyle{ \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}} }[/math] || [math]\displaystyle{ \frac{1}{2}\ln\{(2\pi e)^{N} \det(\Sigma)\} }[/math]||[math]\displaystyle{ \mathbb{R}^N }[/math]

F _ x (vec { x }) = </math > < br/> < math > frac { exp left (- frac {1}{2}(vec { x }-vec { mu }) ^ top Sigma ^ {-1} cdot (vec { x }-vec { mu }) right)}{(2 pi) ^ { N/2}左 Sigma | right | ^ {1/2} < | < math > | < < | < math > frac {1}{ ln (2 pi e){{ n } | math < | | | | > 数学 < bb >

Relation to estimator error

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The differential entropy yields a lower bound on the expected squared error of an estimator. For any random variable [math]\displaystyle{ X }[/math] and estimator [math]\displaystyle{ \widehat{X} }[/math] the following holds:

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[math]\displaystyle{ \operatorname{E}[(X - \widehat{X})^2] \ge \frac{1}{2\pi e}e^{2h(X)} }[/math]

with equality if and only if [math]\displaystyle{ X }[/math] is a Gaussian random variable and [math]\displaystyle{ \widehat{X} }[/math] is the mean of [math]\displaystyle{ X }[/math].

Many of the differential entropies are from.

许多熵的差异来自于。

Differential entropies for various distributions

[math]\displaystyle{ H_h=-\sum_i hf(ih)\log (f(ih)) - \sum hf(ih)\log(h). }[/math]

[数学] h =-sum _ i hf (ih) log (f (ih)-sum hf (ih) log (h)

In the table below [math]\displaystyle{ \Gamma(x) = \int_0^{\infty} e^{-t} t^{x-1} dt }[/math] is the gamma function, [math]\displaystyle{ \psi(x) = \frac{d}{dx} \ln\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)} }[/math] is the digamma function, [math]\displaystyle{ B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} }[/math] is the beta function, and γ_E is Euler's constant.

The first term on the right approximates the differential entropy, while the second term is approximately [math]\displaystyle{ -\log(h) }[/math]. Note that this procedure suggests that the entropy in the discrete sense of a continuous random variable should be [math]\displaystyle{ \infty }[/math]. 右边的第一个术语近似于微分熵，而第二个术语近似于 math >-log (h) </math > 。请注意，这个过程表明，连续随机变量的离散意义上的熵应该是“数学”。 Category:Entropy and information 类别: 熵和信息 Category:Statistical randomness 分类: 统计的随机性 This page was moved from wikipedia:en:Differential entropy. Its edit history can be viewed at 微分熵/edithistory

Table of differential entropies

Distribution Name	Probability density function (pdf)	Entropy in nats	Support
Uniform	[math]\displaystyle{ f(x) = \frac{1}{b-a} }[/math]	[math]\displaystyle{ \ln(b - a) \, }[/math]	[math]\displaystyle{ [a,b]\, }[/math]
Normal	[math]\displaystyle{ f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) }[/math]	[math]\displaystyle{ \ln\left(\sigma\sqrt{2\,\pi\,e}\right) }[/math]	[math]\displaystyle{ (-\infty,\infty)\, }[/math]
Exponential	[math]\displaystyle{ f(x) = \lambda \exp\left(-\lambda x\right) }[/math]	[math]\displaystyle{ 1 - \ln \lambda \, }[/math]	[math]\displaystyle{ [0,\infty)\, }[/math]
Rayleigh	[math]\displaystyle{ f(x) = \frac{x}{\sigma^2} \exp\left(-\frac{x^2}{2\sigma^2}\right) }[/math]	[math]\displaystyle{ 1 + \ln \frac{\sigma}{\sqrt{2}} + \frac{\gamma_E}{2} }[/math]	[math]\displaystyle{ [0,\infty)\, }[/math]
Beta	[math]\displaystyle{ f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)} }[/math] for [math]\displaystyle{ 0 \leq x \leq 1 }[/math]	[math]\displaystyle{ \ln B(\alpha,\beta) - (\alpha-1)[\psi(\alpha) - \psi(\alpha +\beta)]\, }[/math] [math]\displaystyle{ - (\beta-1)[\psi(\beta) - \psi(\alpha + \beta)] \, }[/math]	[math]\displaystyle{ [0,1]\, }[/math]
Cauchy	[math]\displaystyle{ f(x) = \frac{\gamma}{\pi} \frac{1}{\gamma^2 + x^2} }[/math]	[math]\displaystyle{ \ln(4\pi\gamma) \, }[/math]	[math]\displaystyle{ (-\infty,\infty)\, }[/math] Category:Information theory 范畴: 信息论