This is exercise 7.7 from [Martin Hairer's Rough Path notes](http://www.hairer.org/notes/RoughPaths.pdf). 



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> 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. $\Bbb{X}\_{s,t}=\int\_P \Xi$
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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.



Rough path theory- Verify that $\mathbb{X}_{s,t}=\int_s^t X_{s,r} \otimes dX_r$
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?


