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This is exercise 7.7 from Martin Hairer's Rough Path notes.

Verify that $\mathbb{X}_{s,t}=\int_s^t X_{s,r} \otimes dX_r$ where the integral is to be interpreted in the sense of (4.22) (I'll define this later), taking $(Y,Y') $ to be $(X,I)$. In fact, check that this holds not only in the limit $|P|\to 0$ but in fact for every fixed $|P|$, i.e. $\Bbb{X}_{s,t}=\int_P \Xi$

Originally, $\mathbb{X}_{s,t} \colon=\int_s^t X_{s,r} \otimes dX_r$ is a definition for a rough path. However if $Y$ with Gubinelli derivative $Y'$ is a controlled rough path, controlled by $X$, we define:

$$\mathbb{Y}_{s,t} \colon=\int_s^t Y_{s,r} \otimes dX_r\colon = \lim\limits_{|P|\to 0} \int_{p} \Xi$$

where:

$$\Xi_{u,v}= Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v} $$

We define the integral:

$$\lim\limits_{|P|\to 0} \int_{p} \Xi\colon = \lim\limits_{|P| \to 0} \sum\limits_{[u,v]\in P} \Xi_{u,v}$$

So we compute:

$$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}+I\otimes I\mathbb{X}_{u,v}$$ $$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}+\mathbb{X}_{u,v}$$ $$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}+\sum\limits_{[u,v]\in P}\mathbb{X}_{u,v}$$ $$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\left(\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}\right)+\mathbb{X}_{s,t}$$

The last equality is because of telescoping series. I am stuck with the first term though. I need to show that $\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}=0$ and I feel like it's something trivial, but I can't see it.

So my question is, how do we establish $\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}=0$?

Edit, I made a rather elementary mistake, the last step is incorrect. I will post an answer soon.

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2 Answers 2

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Actually, I found my mistake. Once we get to:

$$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}+\mathbb{X}_{u,v}$$

Without loss of generality let $X_s=0$, so:

$$X_{u} \otimes X_{u,v}+\mathbb{X}_{u,v}=X_{s,u} \otimes X_{u,v}+\mathbb{X}_{u,v}$$

Then note Chen's relation:

$$X_{u} \otimes X_{u,v}+\mathbb{X}_{u,v}=\Bbb{X}_{s,v}-\Bbb{X}_{s,u}-\Bbb{X}_{u,v}+\mathbb{X}_{u,v}=\Bbb{X}_{s,v}-\Bbb{X}_{s,u}$$

Note this is a telescoping series

$$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\sum\limits_{[u,v]\in P}\Bbb{X}_{s,v}-\Bbb{X}_{s,u}=\Bbb{X}_{s,t}-\Bbb{X}_{s,s}=\Bbb{X}_{s,t}$$

Can someone confirm this is okay?

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  • $\begingroup$ I don't see how you can take wlog $X_s = 0$ as you suggest in your answer. Note that (as mgubi) noted, you are not using the definition of $\Xi$ suggested by (4.22) since (4.22) reads $\int_s^t Y_{s,r} d X_r$ and not $\int_s^t Y_{r} d X_r$ $\endgroup$ Commented Feb 7, 2022 at 18:29
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I would rather set $\Xi_{u,v}=X_{u,s}\otimes X_{u,v}+\mathbb{X}_{u,v}$. From this we have $$ \Xi_{u,v}+\Xi_{v,w}=X_{u,s}\otimes X_{u,v}+\mathbb{X}_{u,v}+X_{v,s}\otimes X_{v,w}+\mathbb{X}_{v,w} $$ [Chen's relation] $$ = X_{u,s}\otimes X_{u,v}+X_{v,s}\otimes X_{v,w}+\mathbb{X}_{u,w}+X_{u,v}\otimes X_{v,w} $$ $$ = X_{u,s}\otimes X_{u,w}+\mathbb{X}_{u,w}=\Xi_{u,w} $$ from which the result follows.

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  • $\begingroup$ It looks like when applying Chen's relation there is a "-" missing in the last term. However, it works out fine if we define $\Xi_{u,v} := X_{s,u} \otimes X_{u,v} + \mathbb{X}_{u,v}$ instead of as it is right now. $\endgroup$ Commented Feb 7, 2022 at 18:29

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