Skip to main content
6 events
when toggle format what by license comment
Mar 18, 2021 at 17:29 comment added Marcus M There might be some useful ideas in this direction in Azais and Wschebor's book "Level sets and extrema of random processes and fields" (which also has several Kac-Rice theorems that would work right out of the box under strong enough assumptions).
Mar 18, 2021 at 16:59 comment added LostStatistician18 Yes that is a good idea! I had a similar idea before in that what seems like a promising path is to condition on the entire path of one of the processes. If the process you condition on has a countable number of zeroes, and the covariance between it and the other process is suitably non-degenerate, then the conditional probability that they share zeroes should always be zero. This would then only rely on the fact that at least one of the paths has a countable number of zeroes, and some non-degeneracy of the covariance. I was not able to make this rigorous though.
Mar 18, 2021 at 16:16 comment added Marcus M Hmm I'm not sure; I'm not very familiar with many examples without these nice regularity properties, so I don't have a good feel for their behavior. There might be an even softer approach, where you could probably (?) write $X(t) = \rho(t) Z_1(t) + \sqrt{1 - \rho(t)^2} Z_2(t)$ and $Y(t) = \rho(t) Z_1(t) + \sqrt{1 - \rho(t)^2} Z_3(t)$ where the three processes $Z_j$ are independent Gaussian processes. The idea would be to condition on $Z_1, Z_2$ and look at the countable collection of zeros of $X$; then since $Z_3$ is non-degenerate the probability $Y$ is zero at those points will be 0.
Mar 18, 2021 at 4:35 comment added LostStatistician18 Awesome! Thanks Marcus! That is exactly what I was looking for. Do you think that continuous differentiability could be relaxed to still achieve the same result?
Mar 18, 2021 at 4:33 vote accept LostStatistician18
Mar 17, 2021 at 3:05 history answered Marcus M CC BY-SA 4.0