# Multivariate Gaussian Definition when Covariance matrix is singular, What's wrong?

Given $$\mathbf{\Sigma} \in \mathbb R^{k \times k}$$ $$\mathbf{u} \in \mathbb R^k$$ The multivariate Gaussian pdf can be determined By definition: $$f(\mathbf{x})=\frac{1}{2\pi^{\frac{-k}{2}}|\Sigma|^{\frac{1}{2}}}e^{\frac{1}{2}(\mathbf{x-u})^T\mathbf{\Sigma}^{-1}(\mathbf{x-u})}$$

The Covariance matrix is only limited to be positive semidefinite.

So it could be singular (Non-invertible)

This will also lead to a zero in the denumerator, and also the$\Sigma^{-1}$ doesn't exist.

What we do in that case to write the joint pdf?

## 1 Answer

Covariance matrix of multivariant Gaussian is square, symmetric (both for all covariance matrices) and positive definite (semi-positive definite for all covariance matrices). So why is the matrix positive definite? It we has one or more dimensions that have variance = 0 this means that they are deterministic (constant) and hence we can neglect these dimensions. If we neglect all dimensions that do not provide any information regarding the RV we will remain with positive definite matrix instead of semi positive definite. A square positive definite matrix is always invertible (det not equal zero).