Infimum of Gaussian process Consider a Gaussian Process $g\sim GP(\mu,k)$ with mean zero $\mu\equiv0$ and continues covariance $k(t_1,t_2)=k(|t_1-t_2|)$ defined on the interval $A=[0,T]$. I'd like to make no assumptions about g(0).
What can be said on the infimum of $g(t)$? For example, what is the distribution of $m=\inf_A g(t)$? What is the mean? What is the median?
Related and perhaps simpler version would be to ask what is $Pr(\{g(t)>c\ :\ \forall t\in A\})=Pr(m>c)$? Or at least what is $Pr(m>0)$, the probability of no zero crossing in $[0,T]$?
Note that unlike related question I'm not interested in the asymptotic behavior for large $c$ but rather what is the probability of infimum near 0 (or how $Pr(m>c)$ behave for small $c$).
EDITED: Per Ylvisaker 1965 we know the expected number of zeros in $A$ is $\frac{T}{\pi}\sqrt{-k''(0)}$. But I couldn't see how it help get $Pr(m>0)$.
 A: Googling "Stationary Gaussian process" and "level crossing" or "first passage" brings up a lot of relevant-looking material. This one seems to address your question most directly out of the ones I peeked at: Zero Crossing Probabilities for Gaussian Stationary Processes by G. F. Newell and M. Rosenblatt. The authors look at the quantity $H_X(T) = P[X(t)>0\text{ for all } 0\le t \le T]$ and establish some upper and lower bounds.
A: I'll summarize what I've learned.
Denote a Gaussian Process $g\sim GP(\mu,k)$ with mean zero $\mu\equiv0$ and continues covariance $k(t_{1},t_{2})=k(|t_{1}-t_{2}|)$ defined on the interval $[0,T]$ and also $m=\inf_{t\in[0,T]}g(t)$. 
I could not find any result regarding the pdf (or cdf) of $m$, not even its mean. Also $P(m>c)$ yielded no results except for the case $c=0$, where from symmetry $P(m>0)=\frac{1}{2}p(0,T)$ with $p(n,T)$ is the probability of $g(t)$ having exactly $n$ zeros in $[0,T]$. This latter quantity is discussed extensively in the literature on “zero crossing intervals” and can be calculated numerically, as we shall now explain.
As already mensioned, the mean number of zeros in time unit is given by Rice' Formula $\beta=\sqrt{-k''(0)}/\pi$ (when $k$ is normalized, $k(0)=1$). Also denote $P_{n}(\tau)$ the probability density for of the event that the $n+1$th zero crossing occuring at interval $\tau$ after the first zero crossing. 
McFadden's paper from 1958, The axis-crossing intervals of random functions--II
show that $U(\tau)=\sum_{0}^{\infty}P_{n}(\tau)$, the probability of some zero crossing at interval $\tau$ after a zero crossing have a (complicated) closed-form expression $U(\tau)=\frac{2}{\pi}(\frac{1}{H(\tau)}-\tan^{-1}H(\tau))R(\tau)$ (eq. 20-21 there) with $R(\tau)=\frac{1}{2\beta\pi}\frac{k(\tau)k'(\tau)^{2}+(1-k(\tau)^{2})k''(\tau)}{(1-k(\tau)^{2})^{3/2}}$ and $H(\tau)=\frac{k(\tau)k'(\tau)^{2}+(1-k(\tau)^{2})k''(\tau)}{\sqrt{[k'(\tau)^{2}+(1-k(\tau)^{2})k''(0)]^{2}-[k(\tau)k'(\tau)^{2}+(1-k(\tau)^{2})k''(\tau)]^{2}}}$. Now $P_{0}(\tau)$ can be calculating numerically from the convolutional equation $P_{0}(\tau)=\frac{1}{2}[U(\tau)+R(\tau)]-\frac{1}{2}[U(\tau)-R(\tau)]*P_{0}(\tau)$ (eq. 47 there). Then $p(0,T)$ can be calculated numerically by integrating $p(0,T)=1+\beta[\int_{0}^{T}d\tau\int_{0}^{\tau}dlP_{0}(l)-\tau]$.
