This question is about Gaussian processes interpreted as distributions over the space of functions. Gaussian processes have the property that their integrals are Gaussian random variables; cf. this StackExchange post for a derivation. In particular, the integral of a zero-mean Gaussian process $X(t)$ over $t$ is almost always nonzero.

Does there exist a "canonical" distribution over functions $X(t)$ over $[0,1]^n$ that integrate to zero (almost always)?

Ideally, it'd be great to identify a distribution that's easy to work with computationally, e.g. one for which we can draw samples of $X(t_k)$ for some fixed/finite set of values $t_k\in[0,1]^n$.

Of course the simplest thing to do would be to subtract the mean, i.e. to take a standard zero-mean Gaussian process $X(t)$ and define a new function $X_{\mathrm{centered}}(t):=X(t)-\mathbb E_{t\in[0,1]}[X(t)]$, but I'm worried that this distribution is hard to work with computationally because it couples different $t$ values together in a way that requires knowing $X(t)$ for all $t$ to evaluate the function.

  • $\begingroup$ en.wikipedia.org/wiki/Brownian_bridge#General_case ? $\endgroup$
    – Glen_b
    Nov 5, 2019 at 2:16
  • $\begingroup$ Hmm, I'm not sure I see it. Seems like the Brownian bridge gives a function that "connects" two values but doesn't have a guarantee on the integral over t. I need that P(a function has nonzero integral)=0, not that the expected integral is zero. Does that make sense? Thanks! $\endgroup$ Nov 6, 2019 at 3:12
  • $\begingroup$ The Brownian Bridge is itself the integral of its increments, is it not? $\endgroup$
    – Glen_b
    Nov 6, 2019 at 3:15
  • $\begingroup$ Afraid I don't know this area well enough to interpret what this means. Is the idea to make a Brownian bridge and then somehow differentiate it in $t$? $\endgroup$ Nov 6, 2019 at 3:16
  • $\begingroup$ Brownian motion / a Weiner process is an example of a Gaussian process; its value at time $t$ is Gaussian and is the "integral" of its increments $dW_s$. A generalized Brownian bridge appears to have all the properties you requested in your question. $\endgroup$
    – Glen_b
    Nov 6, 2019 at 3:39

1 Answer 1


Finding the distribution and sampling from it should be straightforward for continuous covariance and mean functions which have readily computable integrals.

Let us assume for convenience that $X$ is defined on $[0;1]$, zero-mean with continuous covariance function $K$. It should not be very difficult to generalise from this. Define $Y=\int_0^1X(s)ds$ as your integrated process.

Existence and properties of $Y$

If $$\int_0^1 \int_0^1 K(s,t)dsdt < \infty$$ then $Y$ is a constant Gaussian process, i.e. a Gaussian random variable. This random variable (not the process) is zero mean and has covariance $C=\int_0^1 \int_0^1 K(s,t)dsdt$.

Joint properties of $(X,Y)$

$X$ and $Y$ are joint Gaussian processes, i.e. they have a joint Gaussian distribution. Their cross covariance function is $$ Cov(X(t),Y) = \int_0^1 K(s,t)ds.$$

Conditioning on Y

As it is true for any Gaussian process, conditioning $(X,Y)$ on $Y$ will give you a Gaussian process again. So define the new Gaussian process as $$Z = X | Y=0$$

Sampling from $Z$

Using the conditioning formula for Gaussian processes you can draw samples from $Z$ in pretty much the same way as you would draw samples from $X$.

A paper describing this in more depth and generality is "Linear Operators and Stochastic Partial Differential Equations in Gaussian Process Regression" by Simo Särkkä. Available online here.

  • $\begingroup$ This is a nice idea! Seems to be a reasonable thing to do for our problem --- thanks so much. $\endgroup$ Nov 19, 2019 at 15:21

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