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.

  • 2
    $\begingroup$ 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. $\endgroup$ Commented Aug 17, 2013 at 20:47
  • $\begingroup$ @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. $\endgroup$
    – caburke
    Commented Aug 20, 2013 at 17:01

2 Answers 2


In the paper Ghosal, Subhashis; Roy, Anindya, Posterior consistency of Gaussian process prior for nonparametric binary regression, Ann. Stat. 34, No. 5, 2413-2429 (2006). ZBL1106.62039., part of the result of Theorem 5 states that

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.


Does the following publication help?

Sudipto Banerjee, Alan E Gelfand & C. F Sirmans (2003), Directional Rates of Change Under Spatial Process Models, Journal of the American Statistical Association, 98:464, 946-954, DOI: 10.1198/C16214503000000909

Section 2 states that for the directional derivatives of a process to exist, it must be mean square differentiable:

[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 $h \rightarrow 0$.


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