Let $X,Y \in C^{\alpha}([0,1],\mathbb{R})$ be $\alpha$-Hölder paths in $\mathbb{R}$ with $1/3 < \alpha < 1/2$. Then a path $Y' \in C^{\alpha}([0,1],\mathbb{R})$ is called a Gubinelli derivative of $Y$ with respect to $X$ if there exists a $C > 0$ s.t.

$$ \vert Y_t - Y_s - Y_s' (X_t - X_s) \vert \leq C \vert t - s \vert^{2 \alpha}, ~~~ \forall s,t \in [0,1]. $$

I am fine with the definition of Gubinelli derivatives, but I am struggling to compute it even for very simple functions. For example, let $X_t := t^{\alpha}$ and $Y_t := t^{\beta}$ with $1/3 \leq \alpha \leq \beta \leq 1/2$. What would be the Gubinelli derivative in this example?

Of course if $\alpha = \beta$, then $Y_s = 1$ for every $s \in [0,1]$ is sufficient.

I also know that once a Gubinelli derivative $Y'$ is found, then for a sufficiently regular function (say $\phi \in C^2_b$) one can find $(\phi(Y))'$ explicitely.


1 Answer 1


As mentioned in the section 6.2 in "A course in rough paths"-CRP for short, if you can find $t_{n}\to s$ such that

$$\frac{|X_{s,t_{n}}|}{|t_{n}-s|^{2\alpha}}\to +\infty$$

then the Gubinelli-derivative $Y'$ is uniquely determined (Proposition 6.4 in CRP). They have a formula for it:

$$Y'_{s}=\lim_{t_{n}\to s}\frac{Y_{t_{n}}-Y_{s}}{X_{t_{n}}-X_{s}}\label{1}\tag{1}$$

for a particular sequence $t_{n}\to s$.

So working in reverse, if you can show that the above expression $Y'_{s}\in C^{\alpha}$, then you got your Gubinelli-derivative and thus you got a rough-integral formulation (see Theorem 4.10 in CRP).

In the particular case of $X_{t}=t^{\alpha}$, the rough path lift is (Remark 2.2. in CRP)


It is not truly rough because the limit goes to zero:

$$\frac{|X_{s,t_{n}}|}{|t_{n}-s|^{2\alpha}}\to 0.$$

So uniqueness seems unlikely. Now, using the remainder we see for $t\approx s$ and $s$ away from zero:

$$ \frac{\vert Y_t - Y_s - Y_s' (X_t - X_s)\vert}{ \vert t - s \vert^{2 \alpha}}\approx \frac{ s^{\beta-1} \vert t-s\vert -Y'_{s} s^{\alpha-1} \vert t-s\vert}{ \vert t - s \vert^{2 \alpha}}\to 0$$

since $\alpha<\frac{1}{2}$. So any function $Y'_{s}\in C^{\alpha}$ will do eg. $Y'_{s}=s^{\alpha}$ for s away from zero.

However for $s$ close to zero or equal, we get $t^{\beta-2\alpha}-Y'_{0}t^{-\alpha}$ which diverges as $t\to 0$ even if we set $Y'=0$ because $2\alpha>\frac{2}{3}>\frac{1}{2}>\beta$.

If you want me to add more details, please let me know.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .