Let $(X(t))\_{t \in [-1,1]}$ be a centered non-stationary smooth gaussian process with covariation function $\rho(t,s) = \mathbb E[X(t)X(s)]$. For $t\_0 \in (-1,1)$ and $\epsilon \in (-1-t\_0,1-t\_0)$, define
$$
p\_X(t\_0,\epsilon) : = \mathbb P(X(t) = 0\,\text{ for some } t \in [t\_0-\epsilon,t\_0+\epsilon])
$$



> 
> **Question.** *What is a good **upper-bound** for $p\_X(t\_0,\epsilon)$ which is valid for **small** $\epsilon$ (i.e for $\epsilon \to 0^+$) ?*
> 
> 
> 


**A concrete example.** The GP I have in mind is $X(t) := tU + (1-t^2)^{1/2}V$, where $(U,V) \sim N(0,I\_2)$, for which the covariation function is
$$
\rho(t,s) = ts +(1-t^2)^{1/2}(1-s^2)^{1/2}.
$$



Lower-bound on zero-crossing probability of the nonstationary gaussian process $X(t) = tU+(1-t^2)^{1/2}V$, with $(U,V) \sim N(0,I_2)$
$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example,
$$p\_X(t\_0,\ep)=P\Big(m\_1<\frac VU<m\_2\Big),$$
where
$$m\_1:=\min\_{t\in[t\_0-\ep,t\_0+\ep]}r(t)
=r(t\_0+\ep),\quad 
m\_2:=\max\_{t\in[t\_0-\ep,t\_0+\ep]}r(t)
=r(t\_0-\ep),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$
this follows because $r$ is a continuous decreasing function on the interval $(-1,1)$.


Letting now
$$\th\_j:=\arctan m\_j$$
and using the rotational symmetry of the distribution of $(U,V)$, we get
$$p\_X(t\_0,\ep)=\frac{\th\_2-\th\_1}{\pi}.
$$


