2
$\begingroup$

Let $f:\mathbb{R}^d\to\mathbb{R}$ be a smooth compactly supported covariance function of a stationary random fields (hence positive definite).

Is there a compactly supported function $g:\mathbb{R}^d\to \mathbb{C}$ such that $g\ast g=f$ ?

Using Fourier transform and Bochner's theorem, this question is equivalent to the construction of a entire square root for entire function of exponential type that are positive and fast decaying on the real line.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ And how do you expect to get an entire square root when the Fourier transform has simple complex zeroes (which isn't hard to arrange: take any positive and fast decaying on the line entire function of exponential type and multiply it by $z^2+\mu^2$ with some real $\mu$ such that $i\mu$ is not a zero of the original function)? $\endgroup$
    – fedja
    Commented Sep 6, 2020 at 4:55
  • $\begingroup$ Very interesting, I did not think about multiplication by such function to obtain simple zeroes. So indeed, it is not always possible. I just need to make this assumption then. I wonder if it says something interesting about the covariances functions that cannot be root squared that way. $\endgroup$
    – Chris
    Commented Sep 6, 2020 at 8:04

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .