Distribution over functions that integrate to 0

Question

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.

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.