Let $X\_{t}=\sum\_{i=1}^n(1+s\cdot w)\sin(t\_i)$ where $t\in T=[-\pi/2,\pi/2]^n/\{\vec 0\}$, $w\sim\mathbb{N}(0,1)$, $s$ 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 $n\rightarrow\infty$,


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



the infimum of a random process
We have $X\_t=Y\sum\_{i=1}^n\sin(t\_i)$, where $Y:=1+sw$,
$t=(t-1,\dots,t\_n)\in T^n:=[-\pi/2,\pi/2]^n/\{\vec 0\}$, $w\sim N(0,1)$, and $s$ is a real number.


So,
$$\inf\_{t\in T^n}X\_t=\min\_{t\in T^n}X\_t=-n|Y|;$$
this minimum is attained at $t=(-\frac\pi2,\dots,-\frac\pi2)$ if $Y\ge0$ and at $t=(\frac\pi2,\dots,\frac\pi2)$ if $Y<0$.


If $s=0$, then $Y=1$. If $s\ne0$, then $Y$ is normal random variable. So, in any case, $P(Y\ne0)=1$.


So, for any real $s$,
$$P(\inf\_{t\in T^n}X\_t>0)=0\not\to1.$$


