When fitting a Gaussian process, we use a kernel function to define the covariance matrix. It is well known that if $k(x, y)$ is a kernel function, then $k_1(x, y) = k(x, y)^p$ is also a kernel function for any integer $p$. This is obviously just a result of induction from the fact that multiplication of two kernels is still a kernel.

Is this still true when $p$ is not integer, say $\frac{1}{2}$?

PS: This is motivated by the fact that the arithmetic mean of kernels is still kernels, but sometimes we may want geometric mean instead. If $p$ were allowed to be non-integer, then one could conclude that the weighted geometric mean of kernels is also a kernel, which is very useful in practice.


1 Answer 1


You have exactly defined the class of infinitely divisible kernels, i.e., a kernel $k(x, y)$ such that $k(x, y)^p$ is a kernel for any $p > 0$.

Not all kernels are infinitely divisible. Many of the kernels you know and love are infinitely divisible.

  • $\begingroup$ Could you please add a reference for this topic? $\endgroup$
    – Yves
    Mar 8, 2023 at 9:48
  • $\begingroup$ This link should go directly to a freely downloadable paper google.com/… $\endgroup$ Mar 8, 2023 at 16:50

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