Lower-bound on zero-crossing probability of the nonstationary gaussian process $X(t) = tU+(1-t^2)^{1/2}V$, with $(U,V) \sim N(0,I_2)$

Question

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 $$ \rho(t,s) = ts +(1-t^2)^{1/2}(1-s^2)^{1/2}. $$

Answer 1

$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$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 $$p_X(t_0,\ep)=\frac{\th_2-\th_1}{\pi}. $$

Answer 2

Disclaimer. This is just to push the accepted answer a bit further and obtain an explicit upper-bound, valid for small $\epsilon$.

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As shown by user Iosif, $M:=U/V$ has Cauchy distribution with CDF $F_M(m) := \mathbb P(M \le m)$ given by $$ F_M(m) = \dfrac{1}{2}+\dfrac{\arctan(m)}{\pi}. $$

Thus, if the $m_j$'s are as defined in user Iosif's answer, we ge $$ \begin{split} p_X(t_0,\epsilon) = \mathbb P(m_1 < M < m_2) = F_M(m_2)-F_M(m_1) &= \dfrac{\arctan(m_2)-\arctan(m_1)}{\pi}\\ & \le \dfrac{m_2-m_1}{\pi}, \end{split} $$ where we have used the fact that $\arctan$ is nondecreasing and $1$-Lipschitz continuous.

Now, fix $\alpha \in (0,1)$ and define $C_\alpha := 1/(1-\alpha^2)^{3/2} < \infty$.

The derivative of $r:t \mapsto -t/(1-t)^{1/2}$ is $r'(t) = -1/(1-t^2)^{3/2}$ (for $|t| < 1$), and so $r$ is $C_\alpha$-Lipschitz on the interval $[-\alpha,\alpha]$. We deduce that $m_2-m_1 \le 2C_\alpha\epsilon$.

Thus, for any fixed $\alpha \in (0,1)$ and $\epsilon \to 0^+$, we have the following upper-bound $$ \sup_{-\alpha < t_0 < \alpha}p_X(t_0,\epsilon) \le \frac{2C_\alpha\epsilon}{\pi}=\mathcal O(\epsilon). $$