# Distribution over functions that integrate to 0

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.

• Nov 5, 2019 at 2:16
• 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! Nov 6, 2019 at 3:12
• The Brownian Bridge is itself the integral of its increments, is it not? Nov 6, 2019 at 3:15
• 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 $t$? Nov 6, 2019 at 3:16
• 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. Nov 6, 2019 at 3:39

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.

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