Zeros of a non-degenerate bivariate Gaussian Process

Question

Suppose $(X(t),Y(t))$ $t\in[0,1]$ is a bivariate Gaussian process. We can assume that each component is continuously differentiable, but not necessarily stationary, and that the covariance kernels of $X$, $Y$ and the correlation function between $X$ and $Y$ are non-degenerate in the sense that $C_X(t,t)>0$, $C_Y(t,t)>0$ and $-1 < \rho_{XY}(t,t) <1$ for all $t\in [0,1]$.

My question is: is $P( \{ (X(t),Y(t) ) = 0 \mbox{ for some $t \in [0,1]$ }\})=0$? If so is there a simple proof?

Under the assumption that each component process is continuously differentiable, the number of zeros of each process must be countable, and so intuitively this is true since at the points where either process is zero, the probability that each component process is equal should be zero (since their distribution/covariance is non degenerate). I cannot turn this into a rigorous proof though, and perhaps the conjecture is even incorrect!

Answer 1

This is definitely true under some assumptions about the regularity of $C_X, C_Y$ and $\rho_{XY}$. The easiest way to do this is to find a version of the Kac-Rice formula that applies in your case (you may need slightly more assumptions on regularity).

You could also do it more naively. Assuming $C_X(t,t)$, $C_Y(t,t)$ and $\rho_{XY}(t,t)$ are continuous functions of $t$ for $t \in [0,1]$, then this is true (it is likely that these regularity properties follows from assuming the paths are $C^1$, but the canned facts about that that I know are for stationary processes).

Since $X'$ and $Y'$ are a.s. bounded Gaussian processes, we can set $M$ so that $E \max_t |X'(t)|,E \max_t |Y'(t)| < M$. Also, normalize the processes so that $Var(X(t)) = Var(Y(t)) = 1$ for all $t$; note that by continuity of $C_X,C_Y$, this will only change $M$ by a bounded factor.

Now, at each point $t \in [0,1]$, we have that $(X(t),Y(t))$ is Gaussian whose covariance matrix has $1$'s on the diagonal, and whose off diagonal entry is uniformly bounded away from $1$. Thus, there is a constant $C$ so that for all $t \in [0,1]$ we have $$P(|X(t)| < \epsilon, |Y(t)| < \epsilon) < C \epsilon^2\,.$$

Let $n$ be large (we will take it to infinity) and set $t_j = j/n$. Then $$P(\bigcup_{j} |X(t_j)|< n^{-2/3} \text{ and } |Y(t_j)| < n^{-2/3}) \leq C n \cdot n^{-4/3} = C n^{-1/3}\,.$$

Further, we have $$P(\max_t |X'(t)| > n^{1/3} \text{ or }\max_t |Y'(t)| > n^{-1/3}) \leq 2M n^{-1/3}\,.$$

Conditioned on the event that both derivatives are at most $n^{1/3}$, then if there is a point $t$ with $X(t) = Y(t) = 0$, we must have $|X(t_j)| \leq n^{-2/3}$ and $|Y(t_j)| \leq n^{-2/3}$ for some $j$. This shows $$P(\exists~t \in [0,1] \text{s.t.} X(t) = Y(t) = 0) \leq n^{-1/3}\,.$$

Taking $n \to \infty$ does the trick.