find the coefficient of the multivariate normal distribution?

Question

We know that the multivariate normal distribution is given by $$f(x)=\frac 1 {c} e^{-\frac 1 2(x-\mu)^T\Sigma (x-\mu)}$$ Where $c =\sqrt {\det(\Sigma)2\pi}$

How do we derive this value for $c$?

EDIT: I understand that $c=\int e^{-\frac 1 2(x-\mu)^T\Sigma (x-\mu)dx}$ (which is a multivariable integral), but I don't know how to calculate that.

Answer 1

Basically it has to do with the Jacobian transformation of $X$, which correspond to the eigen value decomposition of $\Sigma$.

$\Sigma=Q\Lambda Q'$

$Q$ is orthogonal, and $\Lambda$ is diagonal. Then we also have

$Y=Q'X, Y\sim N(Q'\mu,\Lambda) $

$Y_i$ are mutually independent, and each has variance $\Lambda_{ii}$. So the constant in its pdf is $\sqrt{\prod{2\pi\Lambda_{ii}}}=\sqrt{(2\pi)^pdet(\Lambda)}=\sqrt{(2\pi)^pdet(\Sigma)}$

Answer 2

The pdf for the Normal distribution is defined as:

$$f(x)={1\over{\sqrt{2 \pi\sigma^2}}}e^{-(x-\mu)^2\over{2 \sigma^2}}$$

For the multivariate normal distribution instead of using a single value for variance, we are given a Covariance Matrix $\Sigma$, therefore in our case (where $\Sigma$ is assumed to be positive definite) we are looking to take the determinant of $\Sigma$ which will give us the generalized variance¹.

I hope this insightful!

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_{¹The generalized variance (GV) of a p-dimensional random vector variable X is de- fined as the determinant of its variance-covariance matrix. GV was introduced by Wilks as a scalar measure of overall multidimensional scatter (Source)}