Let $X\_{t}=\sum\_{i=1}^n(1+s\cdot w\_i)t\_i\sin(t\_i)$ where $t\in T=[-\pi/2,\pi/2]^n/\{\vec 0\}$, $w\_i$ are iid standard gaussian variables, $s$ is a scalar denoting the strength of Gaussian noise.


How to find the condition on $s$ such that $X\_t$ is strictly positive with high probability? i.e. when $n\rightarrow\infty$,


$$P(\inf X\_t>0)\rightarrow 1.$$


**What I tried:**
I tried to prove $P(X\_t>0)\rightarrow 1$ for every $t\in T$. This can be computed exactly by using the symmetric property of gaussian and gaussian tail bound. However this gives a unsatisfactory bound on $s$. Thus I am seeking for 'epsilon-net' type idea as follows.


I've already found covering consitituted by uniformly sampled points in the region $T$ as the centers of the balls with $\epsilon$ radius. Also, on the random centers, I prove that $P(X\_t>0)\rightarrow 1$ where $t\in N$ and $s<\sqrt{n}$. This nice bound in $s$ gives me a hope that it also apply to the whole region $T$ instead of only sampled points. Thus I came up with this epsilon net argument to deal with the points that are not on the net
\begin{equation}
\begin{aligned}
P(\inf\_{t\in T} X\_t>0)&=1-P(\inf\_{t\in T} X\_t<0)\\&=1-P(\inf\_{t\in T} (X\_t-X\_{\pi(t)}+X\_{\pi(t)})<0)\\&\geq 1-P(\inf\_{t\in T} (X\_t-X\_{\pi(t)})<0)+P(\inf X\_{\pi(t)}<0)\\
\end{aligned}
\end{equation}


The second probability term $P(\inf X\_{\pi(t)}<0)$ can be lower bounded by union bound.


However, I was stuck on bounding the first probability term.


Normally what we did is $P(\sup\_{t\in T} X\_t-X\_{\pi(t)}>0)\leq P(\sup\_{t\in T}|X\_t-X\_{\pi(t)}|>0)$ and assuming the process is Lipschitz, i.e. $|X\_t-X\_{\pi(t)}|\leq C|t-\pi(t)|$ where $C$ is a random variable, then the probability can be further bounded by $P(C\epsilon>0)$.


Following the same procedure here, assuming we have $|X\_t-X\_{\pi(t)}|<C|t-\pi(t)|$. Then we have the lower bound of the first term $$P(\inf\_{t\in T} (X\_t-X\_{\pi(t)})<0)\leq P(\inf\_{t\in T} C|t-\pi(t)|>0).$$


Clearly, this bound is $1$. Then we would have no chance to prove the first probability goes to $0$ when $n\rightarrow \infty$. I think the issue is that we lost too much when doing $X\_t-X\_{\pi(t)}<C|t-\pi(t)|$.


Following [this book](https://web.math.princeton.edu/%7Ervan/APC550.pdf), I also tried to look into chaining (dealing with probabilistic Lipschitz bound) and slicing (dealing with $E(X\_t)\neq 0$ case). But I think the issue can not be resolved by applying these more technical methods.


This problem looks quite the same as what epsilon net method, chaining trying to prove. But it seems that they do not apply and I was not able to see what exactly the issue is. Is it because I am actually asking a totally different question than the common one, i.e. $P(\sup X\_t>x)<e^{-x^2}$ where $x>0$?



Lower bounding the infimum of a random process
This will not hold under any condition on $s$.


Indeed, by the continuity of $X\_t$ in $t$,
$$\inf\_{t\in T} X\_t=\inf\_{t\in[-\pi/2,\pi/2]^n} X\_t
\le X\_{\vec 0}=0.$$
So, for any real $s$,
$$P(\inf X\_t>0)=0\not\to1.$$


