5
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.