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For a one-dimensional x,


This can be shown through integration by parts. There is a good derivation of that here.

I'm wondering, does the same exist for a multivariate Gaussian $$\ p(x|\mu,\Sigma)=\left(2\pi\right)^{-\frac{N}{2}}|\Sigma|^{-\frac{1}{2}}\exp\left(-\frac{1}{2}(x-\mu)^{\textrm{T}}\Sigma^{-1}(x-\mu)\right)$$

Where $\Sigma$ (positive-definite) is the variance and $\mu$ is the mean. Here, I'm using $N$ for the dimension of $x$. That is, does



My first thought is that the identity should be identical because after carrying out the inner products "upstairs" in the exponential and "downstairs" in the coefficient, it becomes a positive scalar.

This calculation is involved in determining the differential entropy of a multivariate Gaussian. For my answer to agree with the answer given in theorem 9.4.1 here, there should be an additional factor of N on the right hand side.

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1 Answer 1

We can compute this directly by a linear change of variables, replacing $\Sigma^{-1/2}(x-\mu)$ by $x$:

\begin{align*} &\int (2\pi)^{-\frac N2}|\Sigma|^{-\frac12}(x-\mu)^T\Sigma^{-1}(x-\mu)\text{exp}\left(-\frac12(x-\mu)^T\Sigma^{-1}(x-\mu)\right)\ dx \\ &= \int (2\pi)^{-\frac N2}x^Tx\exp\left(-\frac12 x^Tx\right)\ dx \\ &= \int (2\pi)^{-\frac N2}(x_1^2+\cdots+x_N^2)\exp\left(-\frac12(x_1^2+\cdots+x_N^2)\right)\ dx \\ \end{align*}

Distributing, we can express this as a sum of $N$ integrals, each of which are equal to

\begin{align*} &\int (2\pi)^{-\frac N2}x_1^2\exp\left(-\frac12(x_1^2+\cdots+x_N^2)\right)\ dx \\ &=\int\cdots\int (2\pi)^{-\frac N2}x_1^2\exp\left(-\frac12x_1^2\right)\cdots\left(-\frac12x_N^2\right)\ dx_1\cdots dx_N \\ &=\left(\int (2\pi)^{-\frac 12}x_1^2\exp\left(-\frac12x_1^2\right)\ dx_1\right) \left(\int (2\pi)^{-\frac 12}\exp\left(-\frac12x_2^2\right)\ dx_2\right) \cdots \left(\int (2\pi)^{-\frac 12}\exp\left(-\frac12x_N^2\right)\ dx_N\right)\\ &= (1)(1)\cdots(1) = 1 \end{align*}

Therefore, $$\int (2\pi)^{-\frac N2}|\Sigma|^{-\frac12}(x-\mu)^T\Sigma^{-1}(x-\mu)\text{exp}\left(-\frac12(x-\mu)^T\Sigma^{-1}(x-\mu)\right)\ dx = N$$ So yes, to be correct, a factor of $N$ must be inserted to your right-hand side (also, the negative sign on the left side in the first occurence of $-\frac12(x-\mu)^T\Sigma^{-1}(x-\mu)$ should be removed).

Alternate proof: Let $X \sim N(\mu,\Sigma)$ be a multivariate Gaussian random vector. Then the components of $Z=\Sigma^{-1/2}(X-\mu)$ are standard Gaussian, and

\begin{align*} &\int (2\pi)^{-\frac N2}|\Sigma|^{-\frac12}(x-\mu)^T\Sigma^{-1}(x-\mu)\text{exp}\left(-\frac12(x-\mu)^T\Sigma^{-1}(x-\mu)\right)\ dx \\ &= E\left((X-\mu)^T\Sigma^{-1}(X-\mu)\right) \\ &= E\left((\Sigma^{-1/2}(x-\mu))^T(\Sigma^{-1/2}(x-\mu))\right) \\ &= E(Z^TZ) \\ &= E(Z_1^2 + Z_2^2 + \cdots + Z_N^2) \\ &= E(Z_1^2)+E(Z_2^2)+\cdots + E(Z_N^2) \\ &= 1+1+\cdots+1 = N \end{align*}

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