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When using Gaussian processes, the covariance matrix $\mathbf{\Sigma}$ is often defined via a covariance function $K$ as follows $$ \mathbf{\Sigma}_{ij} = K(\underline{x}_i, \underline{x}_j) $$ where $\underline{x}_i, \underline{x}_j$ are coordinates of two points in some space of interest, and belong to some finite set of $n$ such points, resulting in a $n \times n$ covariance matrix.

Is it known whether the Gaussian (or 'squared exponential') covariance function $$ K(\underline{x}_i, \underline{x}_j) = a^2 \exp{\left(-\frac{(\underline{x}_i - \underline{x}_j)^{\top}(\underline{x}_i - \underline{x}_j)}{2l^2}\right)} $$ will always produce a positive-definite $\mathbf{\Sigma}$ for any chosen set of points and values of $a, l$?

Thanks!

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    $\begingroup$ Positive-definite, No; Nonnegative defiinite (sometimes called positive semidefinite) Yes. $\endgroup$
    – Dilip Sarwate
    Commented Oct 14, 2017 at 22:20
  • $\begingroup$ @DilipSarwate oh interesting - I don't suppose you know a source for a proof of that? Thanks $\endgroup$
    – CBowman
    Commented Oct 15, 2017 at 13:06
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    $\begingroup$ It's got nothing to do with Gaussian random variables or processes per se.. All covariance matrices (no matter what the (finite-variance) random variables are) must be positive semi-definite. See this answer of mine to see why this must be so. $\endgroup$
    – Dilip Sarwate
    Commented Oct 15, 2017 at 16:44

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