I am studying Gaussian processes and I found this results, that seems pretty remarkable to me, but yet I can not find anything online about it. I would be very grateful if someone could help me, with a proof or a good reference to read about it.
So if we have $ {f(t), t\in T\subseteq R^2} $, $ C^1 $ centered stationary Gaussian process then why for any fixed t, $ \nabla f(t) $ and $ f(t) $ are jointly continuous (apparently we can just assume this?), but independent?
Thank you very much for your help in advance!
All the relevant details are contained in the one dimensional case, therefore letting $s,t \in \mathbb{R}$ suppose you have $$ \mbox{cov}(f(s),f(t)) = \omega\left( \frac{|s-t|^2}{2} \right), $$ now Gaussian processes are closed under linear transformations, taking the derivative is a linear transformation of your original gaussian processes with new covariance function given by $$ \begin{align*} \mbox{cov}\left(\dot{f}(s) , f(t) \right) &= (s-t) \cdot \omega^{\prime}\left( \frac{|s-t|^2}{2}\right), \\ \mbox{cov}\left(f(s), \dot{f}(t) \right) &= (t-s) \cdot \omega^{\prime}\left( \frac{|s-t|^2}{2}\right), \\ \mbox{cov}\left( \dot{f}(s), \dot{f}(t) \right)&= \frac{\partial^2}{\partial s \partial t}\omega\left(\frac{|s-t|^2}{2} \right). \end{align*} $$ therefore whenever $s=t$ the covariance between $f(t)$ and $\dot{f}(t)$ vanishes and you get the desired independence. I should stress that we can claim independence because they are jointly Gaussian, infact supposing $f(t)$ is a mean zero Gaussian process with covariance function as above then $$ \begin{bmatrix} f(s) \\ f(t) \\ \dot{f}(s) \\ \dot{f}(t) \end{bmatrix} \sim \mathcal{N}\left( \textbf{0} , \begin{bmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{bmatrix} \right) $$ with $$ \begin{align} \Sigma_{11} &= \begin{bmatrix} \mbox{cov}(f(s),f(s)) & \mbox{cov}(f(s),f(t))\\ \mbox{cov}(f(t),f(s)) & \mbox{cov}(f(t),f(t)) \end{bmatrix}\\ \Sigma_{12} &= \begin{bmatrix} \mbox{cov}(f(s),\dot{f}(s)) & \mbox{cov}(f(s) ,\dot{f}(t)) \\ \mbox{cov}(f(t),\dot{f}(s)) & \mbox{cov}(f(t) ,\dot{f}(t)) \end{bmatrix} \\ \Sigma_{22} &= \begin{bmatrix} \mbox{cov}(\dot{f}(s),\dot{f}(s)) & \mbox{cov}(\dot{f}(s),\dot{f}(t))\\ \mbox{cov}(\dot{f}(t),\dot{f}(s)) & \mbox{cov}(\dot{f}(t),\dot{f}(t)) \end{bmatrix} \end{align} $$ The same argument will go through in exactly the same way for higher dimensions just by redefining the covariance function as $\omega(\frac{1}{2}\| \textbf{s} - \textbf{t} \|^2)$.
Here is a reference that contains the necessary details for infinitely differentiable squared exponential kernel Derivative observations in Gaussian Process Models of Dynamic Systems.