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Let $X$ be a semimartingale and $V$ be a finite variation process. Lemma 3 of this page of almostsuremath proves that

$$[X,V]_t=\int_0^t\Delta X_s\,dV_s=\sum_{s\leq t}\Delta X_s\,\Delta V_s$$

This is the proof (let $t_{k,n}=k\,t/n$ denote the $k$-th point of the equispaced partition of $[0,t]$)

$$ \begin{eqnarray} [X,V]_t & = & \lim_{n\rightarrow\infty}\sum_{k=1}^n(X_{t_{k,n}}-X_{t_{k-1,n}})\,(V_{t_{k,n}}-V_{t_{k-1,n}}) \nonumber\\ &=&\lim_{n\rightarrow\infty}\sum_{k=1}^n\int_0^t1_{\{t_{k-1,n}\leq s\leq t_{k,n}\}}(X_{t_{k,n}}-X_{t_{k-1,n}})\,dV_s\nonumber\\ &=&\int_0^t\lim_{n\rightarrow\infty}\sum_{k=1}^n1_{\{t_{k-1,n}\leq s\leq t_{k,n}\}}(X_{t_{k,n}}-X_{t_{k-1,n}})\,dV_s\nonumber\\ &=&\int_0^t\Delta X_s\,dV_s \end{eqnarray} $$

It is said that the third equality follows from bounded convergence theorem (which allows to exchange the limit and the integral). I cannot fully understand why. First, I guess that this is the step in which we use the fact that V has finite variation (I do not see any other place). My other guess is that the FV property of $V$ allows to use the bounded convergence theorem. My question is: how are we exactly using the bounded convergence theorem?

The integral is with respect to $dV_s$, so I a random measure and this troubles me even more (I have never applied the theorem in this situation).

Snoop
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1 Answers1

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My understanding of this matter is as following: note that the argument is pathwise, i.e. $\omega$ is fixed. Semimartingales have càdlàg (rcll) paths which are thus bounded (again, we are saying that each path $(X_s(\omega))_{s\leq t}$ has its own bound $M_t(\omega)\geq 0$) - at the same time, the integrands $|\sum_{1\leq k\leq n}\mathbf{1}_{(t_{k-1,n},t_{k,n}]}(s)(X_{t_{k,n}}(\omega)-X_{t_{k-1,n}}(\omega))|\leq \sup_{s,u\leq t}|X_s(\omega)-X_u(\omega)|\leq 2M_t(\omega)$ so the sequence of integrands is bounded uniformly. Again, with $\omega$ fixed - we can apply bounded convergence in the Lebesgue-Stjeltjes integral wrt $dV_s$.

Snoop
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  • Ok, so the finite-variation of $V$ is used to guarantee that the Lebesgue-Stjeltjes integral is well-defined, right? A final remark could be that it is enough that $X$ is cadlag, not necessarily a semimartingale. – AlmostSureUser Sep 21 '24 at 17:47
  • @AlmostSureUser indeed I agree. I also concur about the fact that only the càdlàg path property of $X$ is involved in this argument. – Snoop Sep 22 '24 at 03:20