Although the question itself can be expressed succinctly, I couldn't come up with a nice self-explanatory title - suggestions are welcome.


I was investigating whether it would be good idea to use Gaussian Process Regression in my application. In the usual use case one conditions a Gaussian process using sample values and that results in a predictive distribution that can be used for a test input. This is a nice approach for making predictions for a single point, however in my application I need to work with intervals. So here comes the question:

Let $K \in \mathbb{R}$ and $f(t)$ be a Gaussian random process with the known mean function $m(t)$ and the known covariance function $c(t_1, t_2)$. What is $\textrm{Pr}\left\{\exists t \in \left[t_1, t_2\right] \text{ s.t. } f(t) \geq K\right\}$?

With words, I am trying to find the probability of the event that the process will exceed a threshold at least at one point during the interval.

I am interested in both analytical and numerical approaches. You are welcome to assume specific forms for $m(t)$ and $c(t_1, t_2)$ (e.g. $m(t) = 0$ and $c(t_1, t_2) = \text{exp}\left(-\frac{1}{2}\left|t_1 - t_2\right|\right)$) to demonstrate an approach.

Note: Edited to fix typos.

  • $\begingroup$ I think you probably meant $c(t_1, t_2) = \exp(-\frac{1}{2}|t_1 - t_2|)$, correct? $\endgroup$
    – cardinal
    Jul 7, 2012 at 22:32
  • $\begingroup$ @cardinal: Of course, fixing it now. Thanks for pointing it out. $\endgroup$ Jul 8, 2012 at 5:25
  • $\begingroup$ I am accepting Nate's answer as it is the best answer the question got in its current status for more than a year. $\endgroup$ Nov 10, 2013 at 6:39

4 Answers 4


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.

  • $\begingroup$ Thanks for the answer and also pointing me to this nice resource. Intuitively one would expect the probability to be an increasing function of $T = |t_2 - t_1|$ and a decreasing function of $K$. The result you quoted definitely confirms the second intuition. However I don't see any explicit reference to $T$ there; it is absorbed in the constants. It is a very important parameter in my application and I need to express the dependence on it more explicitly. $\endgroup$ Jul 12, 2012 at 14:46
  • $\begingroup$ @Mehmet: In the answer I posted, you get the same form for the bound with quantitative information on the constants, if the process has mean zero and is stationary. The (second) constant $C$ in Nate's answer implicitly depends on $T$, as you note. $\endgroup$
    – cardinal
    Jul 13, 2012 at 1:46

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 $$ \mathbb P(\tau_k < t) = 2\cdot\Phi\left(-\frac{|k|}{\sqrt{t}}\right). $$

  • $\begingroup$ Thanks for the answer. It definitely seems like a step in the right direction. The result you provided answers the question for the special case of $t_1 \rightarrow -\infty$. Do you know how to handle the case of $t_1 \in \mathbb{R}$? $\endgroup$ Jul 6, 2012 at 19:13
  • $\begingroup$ This argument is known as the reflexion principle if you want to look it up. $\endgroup$ Jul 7, 2012 at 18:39

In general, we have no implicit formula for each K. However, we can make some useful bounds. In your case, where the process is stationary, I suggest the book "Level set and extrema ..." of J.M. Azais and M. Wschebor.


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