Anyone have the following proof? If a matrix is semi-definite positive and symmetric then it is a covariance matrix.

  • $\begingroup$ a covariance matrix is a matrix that is a positive semi definite matrix. (any) definiteness can be a property of any matrix. whether a covariance matrix obtains a specific type of definiteness depends on the nature of the data the covariances are being computed for $\endgroup$ – develarist Aug 31 '20 at 15:16
  • 1
    $\begingroup$ it may not be symmetric $\endgroup$ – gunes Aug 31 '20 at 15:19
  • 2
    $\begingroup$ One proof is that (assuming it is symmetric) it is the covariance matrix of the (multivariate) Normal distribution it determines, qed. $\endgroup$ – whuber Aug 31 '20 at 15:21

Let $\mathbf{\Sigma}$ be an arbitrary $n \times n$ real symmetric positive semi-definite matrix. Consider the normal random vector $\mathbf{X} \sim \text{N}(\mathbf{0}, \mathbf{\Sigma})$ with density function:

$$p(\mathbf{x}) = (2 \pi)^{-n/2} \det(\mathbf{\Sigma})^{1/2} \exp \Big( -\frac{1}{2} \mathbf{x}^\text{T} \mathbf{\Sigma} \mathbf{x} \Big) \quad \quad \quad \text{for all } \mathbf{x} \in \mathbb{R}^n$$

The variance/covariance matrix for this random vector is:

$$\mathbb{V}(\mathbf{X}) = \mathbf{\Sigma}.$$

(Hat-tip to whuber in the comments for this answer.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.