《机器学习数学基础》第 416 页给出了连续型随机变量的熵的定义，并且在第 417 页以正态分布为例，给出了符合 N ( 0 , σ 2 ) N(0,\sigma^2) N(0,σ2) 的随机变量的熵。
注意：在第 4 次印刷以及之前的版本中，此处有误，具体请阅读勘误表说明
1. 推导（7.6.6）式
假设随机变量服从正态分布 X ∼ N ( μ , σ 2 ) X\sim N(\mu,\sigma^2) X∼N(μ,σ2) （《机器学习数学基础》中是以标准正态分布为例，即 X ∼ N ( 0 , σ 2 ) X\sim N(0,\sigma^2) X∼N(0,σ2) ）。
根据《机器学习数学基础》的（7.6.1）式熵的定义：
H ( X ) = − ∫ f ( x ) log f ( x ) d x (7.6.1) H(X)=-\int f(x)\log f(x)\text{d}x\tag{7.6.1} H(X)=−∫f(x)logf(x)dx(7.6.1)
其中， f ( x ) = 1 2 π σ e − ( x − μ ) 2 2 σ 2 f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} f(x)=2π σ1e−2σ2(x−μ)2 ，是概率密度函数。根据均值的定义，（7.6.1）式可以写成：
H ( X ) = − E [ log f ( x ) ] H(X)=-E[\log f(x)] H(X)=−E[logf(x)]
将 f ( x ) f(x) f(x) 代入上式，可得：
H ( X ) = − E [ log ( 1 2 π σ e − ( x − μ ) 2 2 σ 2 ) ] = − E [ log ( 1 2 π σ ) + log ( e − ( x − μ ) 2 2 σ 2 ) ] = − E [ log ( 1 2 π σ ) ] − E [ log ( e − ( x − μ ) 2 2 σ 2 ) ] = 1 2 log ( 2 π σ 2 ) − E [ − 1 2 σ 2 ( x − μ ) 2 log e ] = 1 2 log ( 2 π σ 2 ) + log e 2 σ 2 E [ ( x − μ ) 2 ] = 1 2 log ( 2 π σ 2 ) + log e 2 σ 2 σ 2 ( ∵ E [ ( x − μ ) 2 ] = σ 2 , 参阅 332 页 ( G 2 ) 式 ) = 1 2 log ( 2 π σ 2 ) + 1 2 log e = 1 2 log ( 2 π e σ 2 ) \begin{split} H(X)&=-E\left[\log(\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}})\right] \\&=-E\left[\log(\frac{1}{\sqrt{2\pi}\sigma})+\log(e^{-\frac{(x-\mu)^2}{2\sigma^2}})\right] \\&=-E\left[\log(\frac{1}{\sqrt{2\pi}\sigma})\right]-E\left[\log(e^{-\frac{(x-\mu)^2}{2\sigma^2}})\right] \\&=\frac{1}{2}\log(2\pi\sigma^2)-E\left[-\frac{1}{2\sigma^2}(x-\mu)^2\log e\right] \\&=\frac{1}{2}\log(2\pi\sigma^2)+\frac{\log e}{2\sigma^2}E\left[(x-\mu)^2\right] \\&=\frac{1}{2}\log(2\pi\sigma^2)+\frac{\log e}{2\sigma^2}\sigma^2\quad(\because E\left[(x-\mu)^2\right]=\sigma^2,参阅 332 页 (G2)式) \\&=\frac{1}{2}\log(2\pi\sigma^2)+\frac{1}{2}\log e \\&=\frac{1}{2}\log(2\pi e\sigma^2) \end{split} H(X)=−E[log(2π σ1e−2σ2(x−μ)2)]=−E[log(2π σ1)+log(e−2σ2(x−μ)2)]=−E[log(2π σ1)]−E[log(e−2σ2(x−μ)2)]=21log(2πσ2)−E[−2σ21(x−μ)2loge]=21log(2πσ2)+2σ2logeE[(x−μ)2]=21log(2πσ2)+2σ2logeσ2(∵E[(x−μ)2]=σ2,参阅332页(G2)式)=21log(2πσ2)+21loge=21log(2πeσ2)
从而得到第 417 页（7.6.6）式。
2. 推导多维正态分布的熵
对于服从正态分布的多维随机变量，《机器学习数学基础》中也假设服从标准正态分布，即 X ∼ N ( 0 , Σ ) \pmb{X}\sim N(0,\pmb{\Sigma}) X∼N(0,Σ) 。此处不失一般性，以 X ∼ N ( μ , Σ ) \pmb{X}\sim N(\mu,\pmb{\Sigma}) X∼N(μ,Σ) 为例进行推导。
注意：《机器学习数学基础》第 417 页是以二维随机变量为例，书中明确指出：不妨假设 X = [ X 1 X 2 ] \pmb{X}=\begin{bmatrix}\pmb{X}_1\\\pmb{X}_2\end{bmatrix} X=[X1X2] ，因此使用的概率密度函数是第 345 页的（5.5.18）式。
下面的推导，则考虑 n n n 维随机变量，即使用 345 页（5.5.19）式的概率密度函数：
f ( X ) = 1 ( 2 π ) n ∣ Σ ∣ exp ( − 1 2 ( X − μ ) T Σ − 1 ( X − μ ) ) f(\pmb{X})=\frac{1}{\sqrt{(2\pi)^n|\pmb{\Sigma}|}}\text{exp}\left(-\frac{1}{2}(\pmb{X}-\pmb{\mu})^{\text{T}}\pmb{\Sigma}^{-1}(\pmb{X}-\pmb{\mu})\right) f(X)=(2π)n∣Σ∣ 1exp(−21(X−μ)TΣ−1(X−μ))
根据熵的定义（第 416 页（7.6.2）式）得：
H ( X ) = − ∫ f ( X ) log ( f ( X ) ) d x = − E [ log N ( μ , Σ ) ] = − E [ log ( ( 2 π ) − n / 2 ∣ Σ ∣ − 1 / 2 exp ( − 1 2 ( X − μ ) T Σ − 1 ( X − μ ) ) ) ] = − E [ − n 2 log ( 2 π ) − 1 2 log ( ∣ Σ ∣ ) + log exp ( − 1 2 ( X − μ ) T Σ − 1 ( X − μ ) ) ] = n 2 log ( 2 π ) + 1 2 log ( ∣ Σ ∣ ) + log e 2 E [ ( X − μ ) T Σ − 1 ( X − μ ) ] \begin{split} H(\pmb{X})&=-\int f(\pmb{X})\log(f(\pmb{X}))\text{d}\pmb{x} \\&=-E\left[\log N(\mu,\pmb{\Sigma})\right] \\&=-E\left[\log\left((2\pi)^{-n/2}|\pmb{\Sigma}|^{-1/2}\text{exp}\left(-\frac{1}{2}(\pmb{X}-\pmb{\mu})^{\text{T}}\pmb{\Sigma}^{-1}(\pmb{X}-\pmb{\mu})\right)\right)\right] \\&=-E\left[-\frac{n}{2}\log(2\pi)-\frac{1}{2}\log(|\pmb{\Sigma}|)+\log\text{exp}\left(-\frac{1}{2}(\pmb{X}-\pmb{\mu})^{\text{T}}\pmb{\Sigma}^{-1}(\pmb{X}-\pmb{\mu})\right)\right] \\&=\frac{n}{2}\log(2\pi)+\frac{1}{2}\log(|\pmb{\Sigma}|)+\frac{\log e}{2}E\left[(\pmb{X}-\pmb{\mu})^{\text{T}}\pmb{\Sigma}^{-1}(\pmb{X}-\pmb{\mu})\right] \end{split} H(X)=−∫f(X)log(f(X))dx=−E[logN(μ,Σ)]=−E[log((2π)−n/2∣Σ∣−1/2exp(−21(X−μ)TΣ−1(X−μ)))]=−E[−2nlog(2π)−21log(∣Σ∣)+logexp(−21(X−μ)TΣ−1(X−μ))]=2nlog(2π)+21log(∣Σ∣)+2logeE[(X−μ)TΣ−1(X−μ)]
下面单独推导： E [ ( X − μ ) T Σ − 1 ( X − μ ) ] E\left[(\pmb{X}-\pmb{\mu})^{\text{T}}\pmb{\Sigma}^{-1}(\pmb{X}-\pmb{\mu})\right] E[(X−μ)TΣ−1(X−μ)] 的值：
E [ ( X − μ ) T Σ − 1 ( X − μ ) ] = E [ tr ( ( X − μ ) T Σ − 1 ( X − μ ) ) ] = E [ tr ( Σ − 1 ( X − μ ) ( X − μ ) T ) ] = tr ( Σ − 1 E [ ( X − μ ) ( X − μ ) T ] ) = tr ( Σ − 1 Σ ) = tr ( I n ) = n \begin{split} E\left[(\pmb{X}-\pmb{\mu})^{\text{T}}\pmb{\Sigma}^{-1}(\pmb{X}-\pmb{\mu})\right]&=E\left[\text{tr}\left((\pmb{X}-\pmb{\mu})^{\text{T}}\pmb{\Sigma}^{-1}(\pmb{X}-\pmb{\mu})\right)\right] \\&=E\left[\text{tr}\left(\pmb{\Sigma}^{-1}(\pmb{X}-\pmb{\mu})(\pmb{X}-\pmb{\mu})^{\text{T}}\right)\right] \\&=\text{tr}\left(\pmb{\Sigma^{-1}}E\left[(\pmb{X}-\pmb{\mu})(\pmb{X}-\pmb{\mu})^{\text{T}}\right]\right) \\&=\text{tr}(\pmb{\Sigma}^{-1}\pmb{\Sigma}) \\&=\text{tr}(\pmb{I}_n) \\&=n \end{split} E[(X−μ)TΣ−1(X−μ)]=E[tr((X−μ)TΣ−1(X−μ))]=E[tr(Σ−1(X−μ)(X−μ)T)]=tr(Σ−1E[(X−μ)(X−μ)T])=tr(Σ−1Σ)=tr(In)=n
所以：
H ( X ) = n 2 log ( 2 π ) + 1 2 log ( ∣ Σ ∣ ) + log e 2 E [ ( X − μ ) T Σ − 1 ( X − μ ) ] = n 2 log ( 2 π ) + 1 2 log ( ∣ Σ ∣ ) + log e 2 n = n 2 ( log ( 2 π ) + log e ) + 1 2 log ( ∣ Σ ∣ ) = n 2 log ( 2 π e ) + 1 2 log ( ∣ Σ ∣ ) \begin{split} H(\pmb{X})&=\frac{n}{2}\log(2\pi)+\frac{1}{2}\log(|\pmb{\Sigma}|)+\frac{\log e}{2}E\left[(\pmb{X}-\pmb{\mu})^{\text{T}}\pmb{\Sigma}^{-1}(\pmb{X}-\pmb{\mu})\right] \\&=\frac{n}{2}\log(2\pi)+\frac{1}{2}\log(|\pmb{\Sigma}|)+\frac{\log e}{2}n \\&=\frac{n}{2}\left(\log(2\pi)+\log e\right)+\frac{1}{2}\log(|\pmb{\Sigma}|) \\&=\frac{n}{2}\log(2\pi e)+\frac{1}{2}\log(|\pmb{\Sigma}|) \end{split} H(X)=2nlog(2π)+21log(∣Σ∣)+2logeE[(X−μ)TΣ−1(X−μ)]=2nlog(2π)+21log(∣Σ∣)+2logen=2n(log(2π)+loge)+21log(∣Σ∣)=2nlog(2πe)+21log(∣Σ∣)
当 n = 2 n=2 n=2 时，即得到《机器学习数学基础》第 417 页推导结果：
H ( X ) = log ( 2 π e ) + 1 2 log ( ∣ Σ ∣ ) = 1 2 log ( ( 2 π e ) 2 ∣ Σ ∣ ) H(\pmb{X})=\log(2\pi e)+\frac{1}{2}\log(|\pmb{\Sigma}|)=\frac{1}{2}\log\left((2\pi e)^2|\pmb{\Sigma|}\right) H(X)=log(2πe)+21log(∣Σ∣)=21log((2πe)2∣Σ∣)
参考资料
[1]. Entropy of the Gaussian[DB/OL]. https://gregorygundersen.com/blog/2020/09/01/gaussian-entropy/ , 2023.6.4
[2]. Entropy and Mutual Information[DB/OL]. https://gtas.unican.es/files/docencia/TICC/apuntes/tema1bwp_0.pdf ,2023.6.4
[3]. Fan Cheng. CS258: Information Theory[DB/OL]. http://qiniu.swarma.org/course/document/lec-7-Differential-Entropy-Part1.pdf , 2023.6.4.
[4]. Keith Conrad. PROBABILITY DISTRIBUTIONS AND MAXIMUM ENTROPY[DB/OL]. https://kconrad.math.uconn.edu/blurbs/analysis/entropypost.pdf, 2023.6.4.