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 at 9:48
  • $\begingroup$ This link should go directly to a freely downloadable paper google.com/… $\endgroup$ Mar 8 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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