# Applying a linear operator to a Gaussian Process results in a Gaussian Process: Proof

In this paper, it is stated without proof or citation that "Differentiation is a linear operation, so the derivative of a Gaussian process remains a Gaussian process". Intuitively, this seems reasonable, as the linear combination of Gaussian random variables is also Gaussian, and this is just an extension to the case where instead of a vector-valued random variable we have a random variable defined on a function space. But I cannot find a source with a proof and the details of a proof elude me.

Proof Outline Let $$x(t)\sim \mathcal{GP}(m(t),k(t, t^\prime))$$ be a Gaussian process with mean function $$m(t)$$ and covariance function $$k(t, t^\prime)$$, and $$\mathcal{L}$$ a linear operator. For any vector $$T=(t_1,...,t_n)$$, let $$x_T=(x(t_1),...,x(t_n))$$. Then $$x_T\sim \mathcal{N}(m_T,k_{T,T})$$. Now consider the stochastic process $$u(t)=\mathcal{L}x(t)$$. It suffices to show that the finite dimensional distributions of $$u(t)$$ are Gaussian, but translating the action of the linear operator on $$x(t)$$ to the finite dimensional case is giving me trouble.

In the case of differentiation, we have $$u(t)=\mathcal{L}x(t)=\frac{dx}{dt}=\lim_ {h\rightarrow 0}\frac{x(t+h)-x(t)}{h}$$. For all $$h>0$$, the random variable $$v(t)=\frac{x(t+h)-x(t)}{h}$$ is normal, and by interchanging integration and the limit, we have

$$\begin{array}{rcl} m_u(t)&=&E\left(\lim\limits_{h\to 0}\frac{x(t+h)-x(t)}{h}\right)\\ &=&\lim\limits_ {h\to 0}E\left( \frac{x(t+h)-x(t)}{h}\right)\\ &=&\lim\limits_ {h\to 0}\frac{m(t+h)-m(t)}{h}\\ &=&m^\prime(t) \end{array}$$

Of course, we need to verify when this interchange is appropriate. Similarly, we can intuit the covariance function of $$u(t)$$ has the form

$$k_u(t,t^\prime)=\frac{\partial^2 x}{\partial t\partial t^\prime }k(t,t^\prime)$$

but I am having a hard time making the leap from finite approximations to the infinite-dimensional case.

Reference Request If there is any textbook or paper that does more than mention this fact in passing, please let me know.

• You need to know something about the process to be sure that the derivative exists. For example, this can't possibly work for Brownian motion. Aug 17, 2013 at 20:47
• @NateEldredge +1 Agreed. It seems that it depends on the differentiability of the mean function and the covariance function. Obviously $\min (s,t)$ is not differentiable. I have read there is a correspondence between covariance funtions and Sobolev spaces, but I know of no good reference for establishing the connection between reproducing kernel Hilbert spaces and Sobolev spaces. That would be helpful as well. Aug 20, 2013 at 17:01

For a Gaussian process η(·) that has differentiable sample paths with mixed partial derivatives up to order α and the successive derivative processes, $$D^w \eta$$(·) are also Gaussian with continuous sample paths. Also, the derivative processes are sub-Gaussian with respect to a constant multiple of the Euclidean distance.
[The process] $Y(s)$ is mean square differentiable at $s_0$ if there exists a vector $\nabla_Y(s_0)$, such that, for any scalar $h$ and any unit vector $\mathbf{u}$, $$Y(s_0 + h \mathbf{u}) = Y(s_0) + h\mathbf{u}^T\nabla_Y(s_0) + r(s_0, h\mathbf{u})$$ where $\frac{r(s_0, h\mathbf{u})}{h} \rightarrow 0$ in the $L_2$ sense as $h \rightarrow 0$.