# Why do we use determinant for multivariate normal distribution?

While learning statistics, I have a question why is the determinant used in the multivariate normal distribution.

When I look for the answer on the internet, so far every answer I looked at was basically saying that it works, so we use that.

But what I want is if there is a mathematical relation between multivariate normal distribution and determinant (volume factor of linear transformation or some other definition).

There was one answer that by using determinant we can make the integral of the density over $$R^{n}$$ equal to $$1$$. This sounds nice, but if there is another intuition, please share it.

• There are two cases: when the determinant of the covariance matrix is $0$ then the distribution is (with probability $1$) restricted to a lower-dimensional subspace of $\mathbb R^n$; when the determinant is non-zero, then it is a measure of the overall dispersion of the distribution, similar to the variance in the one-dimensional case Commented Feb 22, 2020 at 13:19
• The $\det(\Sigma)$ is the volume of the ellipsoid $(x-\mu)^T \Sigma^{-1} (x-\mu)$ that is in the exponential of the multivariate Gaussian distribution. We want the distribution to be normalized to one (because of probability constraint that sums to one), thus when calculating the integral of the distribution (which is the sum of all probabilities) we get the answer to be $\pi^{\frac{n}{2}}\det(\Sigma)$ ($n$ dimension of random vector) and the normalization comes from there. I know as a matter of fact that you can also interpret it with measure concept but i don't know measure theory, xD sorry. Commented Feb 22, 2020 at 13:22
• Try to working with $n$ independent normal r.v.'s $X_i$ then tell me what the joint pdf will look like. Commented Feb 22, 2020 at 21:12

I don't think there's anything more behind the appearance of the determinant of the covariance matrix, $$\ \Sigma\$$, say, beyond the fact that if $$\ \det(\Sigma)\ne 0\$$ then $$\ \int_{\mathbb{R}^n}e^{-\frac{1}{2}(x-\mu)^\top \Sigma^{-1}(x-\mu)}dx = (2\pi)^{\frac{n}{2}}\sqrt{\det(\Sigma)}\$$, so $$\ N=(2\pi)^{-\frac{n}{2}}\det(\Sigma)^ {-\frac{1}{2}}\$$ is the normalising factor for which $$\ N\int_{\mathbb{R}^n}e^{-\frac{1}{2}(x-\mu)^\top \Sigma^{-1}(x-\mu)}dx = 1\$$.
The integral $$\ \int_{\mathbb{R}^n}e^{-\frac{1}{2}(x-\mu)^\top \Sigma^{-1}(x-\mu)}dx\$$ can be evaluated by change of variables from $$\ x\$$ to $$\ y= U(x-\mu)\$$, where $$\ U\$$ is the orthogonal matrix that diagonalises $$\ \Sigma\$$: $$U^\top\Sigma U=\pmatrix{\sigma_1&0&\dots&0\\ 0&\sigma_2&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots& \sigma_n}\ .$$ Because $$\ \Sigma\$$ is a positive definite matrix whenever $$\ \det(\Sigma)\ne 0\$$, $$\ U\$$ always exists , the eigenvalues $$\ \sigma_1, \sigma_2, \dots,\sigma_n\$$ of $$\ \Sigma\$$ are all positive, and $$\ \det(\Sigma)=\sigma_1\sigma_2\dots\sigma_n\$$.
Because orthogonal matrices preserve volume, the change of variables gives \begin{align} \int_{\mathbb{R}^n}e^{-\frac{1}{2}(x-\mu)^\top \Sigma^{-1}(x-\mu)}dx&=\int_{U^{-1}\mathbb{R}^n+\mu}e^{-\frac{1}{2} y^\top U\Sigma^{-1} U^{-1}y}dy\\ &= \int_{\mathbb{R}^n}e^{-\frac{1}{2} y^\top\left(U^\top\Sigma U\right)^{-1}y}dy\\ &= \int_{\mathbb{R}^n}e^{-\frac{y_1^2}{2\sigma_1}- \frac{y_2^2}{2\sigma_2}-\dots-\frac{y_n^2}{2\sigma_n}}dy\\ &=\int_{-\infty}^\infty e^{-\frac{y_1^2}{2\sigma_1}}dy_1 \int_{-\infty}^\infty e^{-\frac{y_2^2}{2\sigma_2}}dy_2\dots\int_{-\infty}^\infty e^{-\frac{y_n^2}{2\sigma_n}}dy_n\\ &=\sqrt{2\pi\sigma_1}\sqrt{2\pi\sigma_2}\dots \sqrt{2\pi\sigma_n}\\ &= (2\pi)^\frac{n}{2}\sqrt{\det(\Sigma)} \end{align}