# Derivatives of Gaussian processes

A Gaussian process $X$ on Euclidean space $\mathbb R^d$ has a radial basis kernel if for any $u,w\in\mathbb R^d$, we have $$\mathrm{Cov}(X_u, X_w) = \sigma^2 \exp\left ( -\frac{\left\lVert u-w \right \rVert^2}2\right)$$

Draws from Gaussian processes with zero mean and radial basis covariance kernels are smooth almost surely. Is there any work on the distribution of the derivative of a draw from such a GP?

• Isn't Ornstein-Uhlenbeck with zero mean a counterexample to the first sentence? Commented Dec 28, 2017 at 7:50
• @AbdelmalekAbdesselam For Brownian motion the covariance of $W_t$ and $W_s$ is $\min(s,t)$. If instead it were a function of $|s-t|$ then it would be a radial basis covariance kernel. Commented Dec 29, 2017 at 6:20
• @BjørnKjos-Hanssen I edited the question to make it clearer, I was referring to the exponentiated squared Euclidean distance, which is often called a radial basis kernel in machine learning. Commented Aug 9, 2018 at 2:27

This should be a good first step: http://mlg.eng.cam.ac.uk/mchutchon/DifferentiatingGPs.pdf

• This is exactly what I was looking for, thank you! Commented Aug 9, 2018 at 2:22

In the following, I assume a zero mean function, for simplicity.

The realizations of a stationary Gaussian process are (a.s.) $$n$$-times differentiable if the covariance function $$k(t_1-t_2):=K(t_1,t_2)$$ is $$2n$$-times differentiable at 0. Hence, in the following, I assume the Gaussian radial basis function

$$k(t) = \exp\left(-\frac{1}{2} t^2\right)$$

as covariance function, which is smooth at zero.

How do the realizations of such a Gaussian process look like? The covariance function $$k$$ is strictly positive, hence any function value is positively correlated with any other function value. However, this correlation is almost zero for long time differences. Hence, such realizations tend to go back of the mean after a short time. In short: what goes up must come down (a.s.).

Now let us have a look at the covariance of the derivative Gaussian process. By the formulas provided by cknoll, its covariance function is

$$k_\partial(t) :=-\frac{\operatorname{d}^2}{\operatorname{d}\!t^2} k(t) = (1-t^2) \exp\left(-\frac{1}{2} t^2\right) = (1-t^2)k(t)$$

and it is easy to see that such processes are again smooth.

So how do the realizations of this derivative Gaussian process look like? For simplicity, I want to distinguish three regions of this covariance function $$k_\partial(t)$$.

1. Close to zero, the covariance function is positive and "looks like" $$k(t)$$. Hence, the derivative Gaussian process locally looks like the original Gaussian process.
2. $$k(\pm1)=0$$ and it is negative for $$|t|>1$$. Hence, derivatives have a negative correlation for time difference $$>1$$. The minimum of $$k_\partial(t)$$ is $$\sqrt3$$. Hence, the original Gaussian process will probably come down again after $$1$$ time steps and the derivative Gaussian process will probably have switched sign after $$\sqrt3$$ time steps.
3. For $$|t|$$ big, the covariance function is almost zero again, hence we would expect it to go back to the mean with a high variance.

In summary, the derivatives of the realizations are again smooth (a.s.), look like the original Gaussian process over very short and very long time scales, but have the tendency to regularly switch their sign over medium time scales.