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Relevant section of article

As of time of writing, the article says:

Suppose $A$ is a symmetric positive-definite (hence invertible) $n \times n$ precision matrix, which is the matrix inverse of the covariance matrix. Then,

$\int_{-\infty}^\infty \exp{\left(-\frac 1 2 \sum\limits_{i,j=1}^{n}A_{ij} x_i x_j \right)} \, d^nx =\int_{-\infty}^\infty \exp{\left(-\frac 1 2 x^{T} A x \right)} \, d^nx=\sqrt{\frac{(2\pi)^n}{\det A}} =\sqrt{\frac{1}{\det (A / 2\pi)}} =\sqrt{\det (2 \pi A^{-1})}$

where the integral is understood to be over $R^n$.

I am wondering if the three determinants on the far right should be $|\mathrm{det}A|,|\mathrm{det}(A/2\pi)|, |\mathrm{det}(2\pi A^{-1})|$ - in other words, have an absolute value sign. Reason for this is if the determinant is negative, then the three expressions are all imaginary, but I don't see why the Gaussian integral would a priori be imaginary or even how it can be imaginary in the first place, given that $A$ is positive-definite.

Allure
  • 774

1 Answers1

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$A$ is positive-definite, so $\det A > 0$, $\det(A/2\pi) > 0$, and $\det(2\pi A^{-1}) > 0$.

See Does a positive definite matrix have positive determinant for more on this.

Eric Towers
  • 70,953