Entropy of Normal Distribution
Theorem
The entropy of the normal distribution $N(\mu, \sigma^{2})$ (when using natural logarithms) is as follows.
$$ H = \dfrac{1}{2} \ln (2\pi e \sigma^{2}) = \ln \sqrt{2\pi e \sigma^{2}} $$
The entropy of the multivariate normal distribution $N_{p}(\boldsymbol{\mu}, \Sigma)$ is as follows.
$$ H = \dfrac{1}{2}\ln \left[ (2 \pi e)^{p} \left| \Sigma \right| \right] = \dfrac{1}{2}\ln (\det (2\pi e \Sigma)) $$
$\left| \Sigma \right|$ is the determinant of the covariance matrix.
Description
The mean $\mu$ does not affect the entropy. The entropy of the standard normal distribution $N(0,1)$ when using natural logarithms is approximately $H = \ln \sqrt{2\pi e } \approx 1.4189385332046727$. Even if a log base $2$ is used, the form of the formula remains the same, and its value is,
$$ H = \log_{2} \sqrt{2\pi e } \approx 2.047095585180641 $$
Proof
Univariate Normal Distribution
To show this, we use that the integral of $p(x) = \dfrac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left( - \dfrac{(x-\mu)^{2}}{2\sigma^{2}} \right)$ is $1$.
$$ \begin{align*} H &= - \int_{-\infty}^{\infty} p(x) \ln p(x) dx \\ &= - \int_{-\infty}^{\infty} p(x) \ln \left[ \dfrac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left( - \dfrac{(x-\mu)^{2}}{2\sigma^{2}} \right) \right] dx \\ &= - \int_{-\infty}^{\infty} p(x) \ln \dfrac{1}{\sqrt{2\pi\sigma^{2}}} dx - \int_{-\infty}^{\infty} p(x) \ln \exp\left( - \dfrac{(x-\mu)^{2}}{2\sigma^{2}} \right) dx \\ &= -\ln \dfrac{1}{\sqrt{2\pi\sigma^{2}}} + \int_{-\infty}^{\infty} p(x) \dfrac{(x-\mu)^{2}}{2\sigma^{2}} dx \\ &= \ln \sqrt{2\pi\sigma^{2}} + \dfrac{1}{2\sigma^{2}}\int_{-\infty}^{\infty} p(x) (x-\mu)^{2} dx \\ &= \ln \sqrt{2\pi\sigma^{2}} + \dfrac{1}{2\sigma^{2}} E[(X-\mu)^{2}] \\ &= \ln \sqrt{2\pi\sigma^{2}} + \dfrac{1}{2\sigma^{2}}\sigma^{2} \\ &= \ln \sqrt{2\pi\sigma^{2}} + \dfrac{1}{2} \\ &= \ln \sqrt{2\pi\sigma^{2}} + \ln \sqrt{e} \\ &= \ln \sqrt{2\pi e \sigma^{2}} \end{align*} $$
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Multivariate Normal Distribution
Since the probability density function of the multivariate normal distribution is $p(\mathbf{x}) = \dfrac{1}{\sqrt{(2\pi)^{p} \left| \Sigma \right|}} \exp \left( -\dfrac{1}{2}(\mathbf{x} - \boldsymbol{\mu})^{T} \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) \right)$,
$$ \begin{align*} H(p) &= -\int p(\mathbf{x}) \ln(p(\mathbf{x}))d \mathbf{x} \\ &= -\int p(\mathbf{x}) \ln \left[ \dfrac{1}{\sqrt{(2\pi)^{p} \left| \Sigma \right|}} \exp \left( -\dfrac{1}{2}(\mathbf{x} - \boldsymbol{\mu})^{T} \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) \right) \right] \\ &= -\int p(\mathbf{x}) \ln\left( \dfrac{1}{\sqrt{(2\pi)^{p} \left| \Sigma \right|}} \right)d \mathbf{x} + \dfrac{1}{2}\int p(\mathbf{x}) (\mathbf{x} - \boldsymbol{\mu})^{T} \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu})d \mathbf{x} \\ &= -\ln\left( \dfrac{1}{\sqrt{(2\pi)^{p} \left| \Sigma \right|}} \right)\int p(\mathbf{x}) d \mathbf{x} + \dfrac{1}{2} E \left[ (\mathbf{x} - \boldsymbol{\mu})^{T} \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) \right] \\ &= -\ln\left( \dfrac{1}{\sqrt{(2\pi)^{p} \left| \Sigma \right|}} \right) + \dfrac{1}{2} E \left[ (\mathbf{x} - \boldsymbol{\mu})^{T} \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) \right] \end{align*} $$
The second term is calculated as follows.
$$ \begin{align*} E \left[ (\mathbf{x} - \boldsymbol{\mu})^{T} \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) \right] &= E \left[ \tr \left( (\mathbf{x} - \boldsymbol{\mu})^{T} \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) \right) \right] \\ &= E \left[ \tr \left( \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) (\mathbf{x} - \boldsymbol{\mu})^{T} \right) \right] \\ &= \tr \left[ E \left( \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) (\mathbf{x} - \boldsymbol{\mu})^{T} \right) \right] \\ &= \tr \left[ \Sigma^{-1} E \left( (\mathbf{x} - \boldsymbol{\mu}) (\mathbf{x} - \boldsymbol{\mu})^{T} \right) \right] \\ &= \tr \left[ \Sigma^{-1} \Sigma \right] \\ &= \tr \left[ I_{p\times p} \right] \\ &= p \end{align*} $$
- The first equality is because of $1 \times 1$ matrix $A$ for $A = \tr(A)$,
- The second equality is due to the cyclic property of trace,
- The third equality is because expectation and trace are exchangeable,
- The fourth equality is due to the properties of expectation of a matrix,
- The fifth equality is due to the definition of the covariance matrix.
Therefore, the entropy is as follows.
$$ \begin{align*} H(p) &= -\ln\left( \dfrac{1}{\sqrt{(2\pi)^{p} \left| \Sigma \right|}} \right) + \dfrac{1}{2} E \left[ (\mathbf{x} - \boldsymbol{\mu})^{T} \Sigma^{-1} (\mathbf{x} - \boldsymbol{\mu}) \right] \\ &= \dfrac{1}{2} \ln \left[ (2\pi)^{p} \left| \Sigma \right| \right] + \dfrac{1}{2}p \\ &= \dfrac{1}{2} \ln \left[ (2\pi)^{p} \left| \Sigma \right| \right] + \dfrac{1}{2}\ln e^{p} \\ &= \dfrac{1}{2} \ln \left[ (2\pi e)^{p} \left| \Sigma \right| \right] \end{align*} $$
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