4
$\begingroup$

I apologise for the confusion of the following sentences. I'm lazy to give more information about Rough path theory as Is a fairly broad subject.

On page 14 of "A Course on Rough Paths With an Introduction to Regularity Structures" by Peter K. Friz & Martin Hairer has written:

For $\alpha \in (1/ 3; 1 /2]$, define the space of $\alpha$-Hölder rough paths (over V ), in symbols $\mathcal C^{\alpha} ([0,T]; V )$, as those pairs $(X; \mathbb X) =: \mathbf{X}$ such that $$ ||X||_{\alpha}:= \sup_{ s\neq t \in [0;T ]} \frac{|X_{s,t}|}{|t-s|^{\alpha}} < \infty , \quad ||X||_{2\alpha}:=\sup_{ s\neq t \in [0;T ]} \frac{|\mathbb X_{s,t}|}{|t-s|^{\alpha}} < \infty , $$ and such that the algebraic Chen relation ( is satisfed.

And on page 56 it hase written: Given a path $X \in \mathcal C^{\alpha}([0, T ]; V )$, we say that $Y \in \mathcal C^{\alpha}([0, T ]; \hat{W} )$, is controlled by $X$ if there exists $Y' \in \mathcal C^{\alpha}([0, T ]; \mathcal L(V , \hat{W})$, so that the remainder term $R^Y$ given implicitly through the relation $$ Y_{s,t }= Y_{s0} X_{s,t} + R_{s,t}^Y , $$ satisfes $||R^Y||_{ 2 \alpha}< 1$. This defines the space of controlled rough paths, $(Y, Y') \in \mathcal D_X^{2α}([0, T ]; \hat{W })$: Although $Y'$ is not, in general, uniquely determined from Y. We call any such $Y'$ the Gubinelli derivative of $Y$ (with respect to $X$). Here, $R_{s,t}^Y$ takes values in $\hat{W}$, and the norm $|| \cdot||$.

Question : What is the intuition behind this idea of the Gubinelli derivative?

Any help is appreciated with thanks in advance.

$\endgroup$
3
$\begingroup$

In a way it is very much like a usual derivative. Recall first that for a regular function $Y$, its derivative $Y'_s$ at a point $s$ is the (unique) number such that $$ Y_{t,s}=Y'_s(t-s)+ R_{s,t}, $$ where $R_{s,t}\to0$ faster than linearly. If $Y$ is twice differentiable, then $R_{s,t}\lesssim |t-s|^2$. That is, as a function of $t$, $Y_t$ "looks like" the linear function $Y_s+Y'_s(t-s)$, in the neighborhood of $s$.

Now simply replace the linear function by $X$. So we impose $$ Y_{t,s}=Y'_sX_{t,s}+R_{s,t} $$ with the remainder $R_{s,t}\to0$ faster than the first term, that is, faster than $|t-s|^\alpha$ (The condition $R_{s,t}\lesssim|t-s|^{2\alpha}$ from Friz-Hairer corresponds to the twice differentiable scenario in the previous case). Then as a function of $t$, $Y_t$ "looks like" the path $Y_s+Y'_sX_{s,t}$. This is great news for integration: we can of course integrate $Y_s$ against $dX_t$ (since as a function of $t$ it is just constant), and we can also integrate $X_{s,t}$ against $dX_t$ (by the definition of a rough path).

Actually, I wouldn't focus so much on assigning a meaning to $Y'$ itself, but rather focus on what the existence of a $Y'$ means for $Y$.

$\endgroup$
2
  • 1
    $\begingroup$ I like this answer very much. With this in mind, in many situations the approximation $Y_{t,s}\simeq Y'_sX_{t,s}$ gives a good idea of what $Y'$ should be: intuitively, if $Y=f(X)$, then $Y_{t,s}\simeq f'(X_s)X_{t,s}$; if $Y=\int A\mathrm dX$ for some process $A$, then $Y_{t,s}\simeq A_s\cdot X_{t,s}$; etc. $\endgroup$ – Pierre PC Jun 3 '20 at 0:12
  • $\begingroup$ Morality: this derivative is none other than the usual derivative modulo the regularities as it has just been pointed out in the commentary of Pierre PC. $\endgroup$ – Furdzik Jun 3 '20 at 10:36
2
$\begingroup$

We want to define $\int_0^T f(X_s) dX_s$ for smooth bounded $f$ with bounded derivatives of all orders. Using linearity and a partition ${t_k}$ of $[0,T]$, we have

\begin{align*}\int_0^T f(X_s) dX_s&=\sum_k\int_{t_k}^{t_{k+1}}f(X_s) dX_s\\&=\sum_k\int_{t_k}^{t_{k+1}}f(X_{t_k})+f'(X_{t_k})(X_s-X_{t_k})+O(|s-t_k|^{2\alpha})dX_s\\&=\sum_k f(X_{t_k})(X_{t_{k+1}}-X_{t_k})+f'(X_{t_k})\int_{t_k}^{t_{k+1}}(X_s-X_{t_k})dX_s+O(|t_{k+1}-t_k|^{3\alpha})\end{align*}

As $3\alpha>1$ the third term goes to zero as the mesh size goes to $0$. The first term is just a Riemann integral. The second term is the "rough path" term. $f'(X_{t_k})$ is the Gubinelli derivative and $\int_{t_k}^{t_{k+1}}(X_s-X_{t_k})dX_s$ is your area process.

$\endgroup$
1
  • $\begingroup$ That is for a smooth function $f$ where the Gubinelli derivative is intuitively seen as a velocity or tangent vector at a point $X_{t_k}$, but what about function with some regularity. Thanks for your answer. $\endgroup$ – Furdzik Jun 2 '20 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.