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

$$m_u(t)=E(\lim_ {h\rightarrow 0}\frac{x(t+h)-x(t)}{h})=\lim_ {h\rightarrow 0}E( \frac{x(t+h)-x(t)}{h})=\lim_ {h\rightarrow 0}\frac{m(t+h)-m(t)}{h}=m^\prime(t)$$

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. – Nate Eldredge Aug 17 '13 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. – caburke Aug 20 '13 at 17:01

[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 $\mathbf{u}$, \begin{equation}Y(s_0 + h \mathbf{u}) = Y(s_0) + h\mathbf{u}^T\nabla_Y(s_0) + r(s_0, h\mathbf{u})\end{equation} where $\frac{r(s_0, h\mathbf{u})}{h} \rightarrow 0$ in the $L_2$ sense as $h \rightarrow 0$.