In this [paper](http://books.nips.cc/papers/files/nips15/AA70.pdf), 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.