Use Kac-Rice to find expected number of zero-crossings of the process $X_t := \alpha |Z_1| - t\|Z\|^2/m$, $Z \sim N(0, I_m)$

56 Views Asked by At

I'm trying (hard) to understand the Kac-Rice theory for computing the expected number of zero-crossings of a random process. To this end, I've identified a simple problem which I know how to solve without Kac-Rice, and I'm interested in recovering the same solution via Kac-Rice. If I can do this (i.e make a link between what I know, and what I'm trying to know), then maybe I understand Kac-Rice...

Related. Part of this endeavour (to understand Kac-Rice theory) can also be found in this question https://mathoverflow.net/q/372292/78539 I've asked.

The test-bed problem

Let $m$ be a large positive integer and let $Z=(Z_1,\ldots,Z_m)$ be a random vector in $\mathbb R^m$ with iid entries from $N(0,1)$. For fixed $\alpha > 0$, consider the random process $(X(t))_{t \in \mathbb R}$ defined by $$ X(t):= \alpha |Z_1| - t\|Z\|^2/m, \tag{1} $$ Note that this process is continuously differentiable w.r.t to time and has derivative given by $X'(t) = -\|Z\|^2/m \le 0$ for all $t$. Thus $X(t)$ decreases over time along every sample path.

Now, for any $t \ge 0$, define $\mu_t := \mathbb E[\#\{s \in [0, t] \mid X(s) \le 0\}]$. A simple computation reveals that $\mu_t$ is the expectation of a Bernoulli randm variable with success rate $p=\mathbb P(\alpha |Z_1| \le t\|Z\|^2/m)$. Thus if $\Phi$ is the standard normal CDF, then have

Exact formula for expected number of zero-crossings. $$ \mu_t = \mathbb P(\alpha|Z_1| \le t\|Z\|^2/m) \to \mathbb P(\alpha |Z_1| \le t) = 2\Phi(t/\alpha)-1, \tag{2} $$ in the limit $m \to \infty$.

Question. How to obtain (2) directly using Kac-Rice theory ?