# Inverse covariance matrix in the Multivariate guassian

The Multivariate guassian is given by:

$$\mathcal{N}(\mathbf{x} ; \boldsymbol{\mu}, \mathbf{Q})=\frac{1}{\sqrt{|2 \pi \mathbf{Q}|}} \exp \left(-\frac{1}{2}(\mathbf{x}-\boldsymbol{\mu})^{T} \mathbf{Q}^{-1}(\mathbf{x}-\boldsymbol{\mu})\right)$$

Where $$Q^{-1}$$ is the inverse of the covariance matrix. I know this is a quadratic form that measures the distance of x from the mean but why is it the inverse of the covariance matrix? Should'nt it be just the covariance matrix?

If there is some data matrix $$A$$, and the columns are mean subtracted then the distance from the origin of $$Ax$$ is: $$=x^TA^TAx= x^TDx$$

where $$D$$ is the covariance matrix. So why does the inverse occur in the guassian? Can someone properly illustrate why inverse is needed? Is it related to the mahalanobis distance?

• See the second "technical comment" in my post at stats.stackexchange.com/questions/62092/…. The answer might become intuitively obvious when you consider the univariate Gaussian (as a special case of the multivariate).
– whuber
May 8, 2020 at 15:03
• Note that the quadratic form in the exponent is known as the Mahalanobis distance, search this site! Oct 19, 2022 at 1:33

If you let $$Q=\sigma^2$$ in your example and take $$x$$ and $$\mu$$ as scalar values, you get the univariate Gaussian PDF.