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I am trying to derive the entropy in multivariate normal distribution. Let $p(X)$, $X\in\Re^{n\times 1}$ to be the probability density function of multivariate normal distribution such that \begin{equation} p(x) = (2\pi)^{-\frac{n}{2}} \det(\Sigma)^{-\frac{1}{2}} e^{-\frac{1}{2}(X-\mu)^\top\Sigma^{-1}(X-\mu)} \end{equation} So the entropy $H$ is defined by \begin{equation} H = -\int^{\infty}_{-\infty} p(X) \log(p(X))dX = \frac{n}{2}\log(2\pi)\int^{\infty}_{-\infty} p(X) dX +\frac{1}{2}\log( \det(\Sigma)) \int^{\infty}_{-\infty} p(X) dX + \frac{1}{2} \int^{\infty}_{-\infty} p(X) (X-\mu)^\top\Sigma^{-1}(X-\mu) dX \end{equation} Because \begin{equation} \frac{dp(X)}{dX} = p(X) (X-\mu)^\top\Sigma^{-1}, \end{equation} we have \begin{equation} H = \frac{n}{2}\log(2\pi) +\frac{1}{2}\log( \det(\Sigma)) + \frac{1}{2} \int^{\infty}_{-\infty} \frac{dp(X)}{dX} (X-\mu) dX \end{equation}

Can anyone know how to solve this integration problem? \begin{equation} \int^{\infty}_{-\infty} \frac{dp(X)}{dX} X dX \end{equation} by using integration by parts, maybe? Many thanks!

Stephen Ge
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  • See https://math.stackexchange.com/questions/2029707 – Hyperplane Sep 07 '19 at 07:40
  • Many thanks, I know I can solve it using the expectation method, I was wondering, is it possible to solve it by integration? – Stephen Ge Sep 07 '19 at 07:45
  • Of course one could do that, for example by transforming into spherical coordinates. $\Sigma^{-1/2}(x-\mu) = z = r\prod_{k=0}^{n-1}\sin(\varphi_k)$. But essentially it will boil down to a derivation of the variance of a normal distribution. – Hyperplane Sep 07 '19 at 07:51

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