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Let $x\mapsto g(x)$ be a centered gaussian random field on $\mathbb R^m$. Let $x_0 \in \mathbb R^n$, and (assuming regularity conditions) consider the gradient-flow

$$ \dot{x}(t) = -\nabla g(x(t)), \;x(0) = x_0. $$

Integrating the above system gives $$ g(x(t)) = g(x_0) + \int_0^t \langle \nabla g(x(t)),\dot{x}(t)\rangle dt = g(x_0)-\int_0^t\|\nabla g(x(t))\|^2 dt. $$

Question 1. Is there a (Kac-)Rice formula for the zero-crossings of $g(x(t)$, namely $\#\{t \in [0, T] \mid g(x(t)) = 0\}$ ?

One of the things I'm interested in are upper-bounds for $g(x(t))$ as a function of $t$.

Question 2. Can one obtain a Borell-TIS bound for the $\sup_{0 \le t \le T} g(x(t))$.

Note. I'm not necessarily looking for a clear-cut answer (though this would be really cool), but general guidelines on how to go about this.

Examples

As a working examples, one could consider the following "simple" fields

  • Linear gaussian random field wherein $g(x) = w^Th(x)$, with $w \sim N(0,I_k)$ and deterministic $h \in \mathcal C^1(\mathbb R^n \to \mathbb R^k)$.
  • Stationary gaussian random fields
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1 Answer 1

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Here's a toy model that is truly linear

$$g(x) =\frac{1}{2}\sum_{i=1}^n \Lambda_i x_i^2, $$

where $\Lambda_i$ are i.i.d. $N(0,1)$ then

$$x(t)= \Big(e^{-t\Lambda_1} x_1(0),\dotsc, e^{-t\Lambda_n} x_n(0)\Big)$$

$$U(t):= g(x(t))=\frac{1}{2}\sum_{i=1}^n \Lambda_ie^{-2t\Lambda_i} x_i(0)^2. $$

Denote by $N_T(U)$ the number of zeros of $U(t)$ on the interval $[0,T]$. Denote by $p_{U(t)}(u)$ the probability density of $U(t)$. Then the Kac-Rice formula state that $\newcommand{\bE}{\mathbb{E}}$

$$\bE\big[\; N_T(U)\;\big]=\int_0^T\bE\big[ \; |U'(t)|\;|\; U(t)=0\;\big] p_{U(t)}(0) dt, $$

where $\bE[-|-]$ denotes the conditional expectation.

Alternatively we have $\newcommand{\bP}{\mathbb{P}}$ $$ \bE\big[\; N_T(U)\;\big]=\underbrace{\bE\big[\; N_T(U)\;|\;g(U(T))>0\;\big]}_{=0}\bP\big[ g(U(T))>0\big]+\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$ $$ =\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$ $$ =\underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))>0,g(U(T))<0\;\big]}_{=1}\bP\big[ g(U(0))>0,g(U(T))<0\big]+ \underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))<0,g(U(T)) <0\;\big]}_{=0}\bP\big[ g(U(0))<0,g(U(T))>0\big] $$ $$ =\bP\big[ g(U(0))>0,g(U(T))<0\big]. $$

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  • $\begingroup$ Thanks for the worked example (upvoted). Can this integral be evaluated explicity to get $E[N_T(U)] =$ some explicit function of $T$ ? I've expected that such a "simple" model would produce an explicit answer down the line :) $\endgroup$
    – dohmatob
    Commented Sep 23, 2020 at 8:12
  • $\begingroup$ What about the limiting case when $n \to \infty$, does anything simplify (e.g via some LLNs / CLT-type argment) ? $\endgroup$
    – dohmatob
    Commented Sep 23, 2020 at 8:18
  • $\begingroup$ Concerning last but one comment, if we consider the even simpler example where $g(x):= \Lambda^Tx$, then $U(t) = g(x(t)) = g(x(0) + t\Lambda) = \Lambda^Tx(0) + t\|\Lambda\|^2$, so that $U'(t) = \|\Lambda\|^2$. Thus the formula would give $E[N_T(U)] = \int_0^TE[\|\Lambda\|^2 \mid \Lambda^Tx(0) + t\|\Lambda\|^2] p_{U(t)}(0)dt$, but I don't quite see how to proceed to get a concrete number. $\endgroup$
    – dohmatob
    Commented Sep 23, 2020 at 8:26
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    $\begingroup$ I don't know how to compute that conditional expectation. Note that $\mathbb{E}[N_T(U)|g(x_0)<0]=0$ and $\lim_{T\to\infty} \mathbb{E}[N_T(U)|g(x_0)>0]=1$ a.s.. $\endgroup$
    – Liviu Nicolaescu
    Commented Sep 23, 2020 at 8:27
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    $\begingroup$ You need to solve for $t$ the equation $$\sum_{i=1}^n\lambda_i e^{-2\lambda_i} x_i^0=0,$$ and view it as a function of $x_i^0,\lambda_i$. Note that this equation has no solution if $\lambda_i >0$ for any $i$. For example if $n=2$, $a,b>0$ the equation $ae^{-at}x-be^{-2bt} y=0$ so $$1-\frac{b}{a}e^{2(a-b)t}\frac{y}{x}=0$$, $$t=\log\left(\frac{xa}{by}\right)/2(a-b).$$ For higher $n$ analyzing $t(\lambda_i,x_j^0)$ is more involved. $\endgroup$
    – Liviu Nicolaescu
    Commented Sep 23, 2020 at 10:20

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