Calculating the probability of an event defined by a condition on a Gaussian random process

Suppose your process is continuous. (By the Kolmogorov-Centsov continuity theorem, a sufficient condition for this is that $c(s,s) + c(t,t) - 2 c(s,t) \le C |s-t|^\gamma$ for some $C, \gamma > 0$.) Then your process induces a Gaussian measure on $C([0,T])$ and a theorem due to Fernique gives an asymptotic result:

Theorem. There exists $C$ and $\epsilon > 0$ such that $$P\left(\sup_{t \in [0,T]} |f(t)| \ge K\right) \le C \exp(- \epsilon K^2).$$

So the probability is very small when $K$ is large.

You can find a proof in these lecture notes. See sections 4.2 and 4.4.

I would recommend obtaining a copy of the text

H. Cramer and M. R. Leadbetter (1967), Stationary and Related Stochastic Processes, John Wiley & Sons, Inc.

which is available as a 2004 Dover reprint, available, e.g., at the link above.

The whole book is a delight, but I think you will be particularly interested in the later chapters, for example,

• Chapter 10: "Crossing" problems and related topics
• Chapter 11: Properties of streams of crossings
• Chapter 12: Limit theorems for crossings
• Chapter 13: Nonstationary normal processes. Curve crossing problems

As a small taste of results from the text:

Lemma: Let $\xi(t)$ be a zero-mean stationary Gaussian process with almost-surely continuous sample paths. Let $C_u(t)$ denote the number of level crossings of the level $u \in \mathbb R$ in the interval $[0,t]$. Then, $$ \mathbb E C_u(t) = \frac{t}{\pi} \sqrt{\frac{\lambda_2}{\lambda_0}} e^{-u^2/2 \lambda_0}\,, $$ where $\lambda_{2n}$, $n \geq 0$ denotes the $2n$th spectral moment $$ \lambda_{2n} = \int_0^\infty \lambda^{2n} \,\mathrm dF(\lambda) $$ and $F$ is the spectral density function.

We can then relate this to the probability of a crossing via the following.

Lemma: Let $\xi(t)$ be a strictly stationary process with continuous one-dimensional distribution and almost-surely continuous sample paths. Suppose that $\mu := \mathbb E C_u(1) < \infty$. Let $q(t)$ denote the probability that at least one crossing occurs in time $t$. Then, $$ q(t) = \mu t + o(t) \,, $$ as $t \to 0$.

In this way we see that the number of level crossings in one sense behaves roughly like a Poisson process with intensity $\mu$.

This only (barely) scratches the surface. The book covers much more and has additional (obviously somewhat dated) references as well. In particular, you may be interested in following classical work, which you may already be familiar with.

S. O. Rice (1945), Mathematical analysis of random noise (broken link), Bell System Tech. J., vol 24, pp. 46–156.

Consider a standard Wiener process $W_t$ with $$ W_0=0,\quad\mathbb E(W_t)=0,\quad \mathbb E(W_t\cdot W_s)=\min(t,s) $$ and let $k\geq0$ be the threshold of interest.

Define the first passage time as $$ \tau_k = \inf\{t\ :\ W_t = k\}. $$ Due to continuity of $W_t$, $$ \mathbb P(W_t > k) = \mathbb P(\{W_t > k\}\cap\{ \tau_k < t\}). $$ Now note that $$ \mathbb P(\{W_t > k\}\cap\{ \tau_k < t\} =\mathbb P(W_t >k\ \mid\ \tau_k < t)\cdot\mathbb P(\tau_k < t) = \frac12\mathbb P(\tau_k < t) $$ due to the symmetrical nature of the Wiener process (intuitively, if the process is already at $k$ at some point of time, it can end up below or above $k$ later on with equal probabilities).

Hence, the probability that threshold $k$ will be crossed before time $t$ is equal to $$ \mathbb P(\tau_k < t) = 2\cdot\mathbb P(W_t > k) = 2\cdot\Phi\left(-\frac{k}{\sqrt{t}}\right),\quad k>0, $$ where $\Phi$ is a c.d.f. of the standard normal distribution. Using the similar argument for $k<0$, we can conclude $$ \mathbb P(\tau_k < t) = 2\cdot\Phi\left(-\frac{|k|}{\sqrt{t}}\right). $$