# Prove that $\left<X\right>_t=\int_0^t G_s^2ds$ where $X_t=\int_0^t F_sds+\int_0^t G_sdW_s$.

I'm trying to prove that $$\left_t=\int_0^t G_s^2ds$$ when $$X_t=\underbrace{\int_0^t F_sds}_{=:A_t}+\underbrace{\int_0^t G_sdW_s}_{=:M_t},$$ where $$\left_t:=\lim_{n\to \infty }\sum_{i=1}^n(X_i-X_{i-1})^2,$$ in some sense ($$L^2$$ or a.s. depending in which sense it exist).

I denote $$X_i:=X_{\frac{i}{n}}$$.

\begin{align*} \sum_{i=1}^n(X_{i}-X_{i-1})^2&=\sum_{i=1}^n(A_{i}-A_{i-1})^2+2\sum_{i=1}^n (A_i-A_{i-1})(M_{i}-M_{i-1})+\sum_{i=1}^n(M_{i}-M_{i-1})^2. \end{align*} Now $$\sum_{i=1}^n(A_{i}-A_{i-1})^2\underset{n\to \infty }{\longrightarrow }0\quad a.s.$$

$$\sum_{i=1}^n(M_{i}-M_{i-1})^2\underset{n\to \infty }{\overset{L^2}{\longrightarrow} }\int_0^t G_s^2ds.$$

Finally $$\sum_{i=1}^n(A_i-A_{i-1})(M_i-M_{i-1})\leq\sqrt{\sum_{i=1}^n(A_i-A_{i-1})^2}\sqrt{\sum_{i=1}^n(M_i-M_{i-1})^2}.$$

Since one sum converges a.s. to 0 and the other one converges to $$\int_0^t G_s^2ds$$ in $$L^2$$ I can't say anything a priori. Indeed, a priori neither $$A_t$$ converges in $$0$$ in $$L^2$$ nor $$M_t\to \int_0^t G_s^2ds$$ in which sense will converges $$\sum_{i=1}^n(X_i -X_{i-1})^2$$ will converges ? The best thing I can say is that there is a subsequence $$\sum_{i=1}^{n_k}(X_{i}-X_{i-1})^2\underset{k\to \infty }{\longrightarrow }\int_0^t G_s^2ds\quad \text{a.s.}.$$

So, how can I conclude ? Maybe $$\sum_{i=1}^n(A_i-A_{i-1})^2\to 0$$ in $$L^2$$ ? I tried to prove that, but I couldn't conclude (unless if $$(A_i)$$ is uniformly bounded by a $$L^2$$ r.v.)

## 1 Answer

Note that $$\sum_{i=1}^n|A_{t_i}-A_{t_{i-1}}||M_{t_i}-M_{t_{i-1}}|\leq \sup_{|s-t| \leq |\Pi|, s,t \in [0,T]} |M_s-M_t| \int_0^T |F(s)| \,ds.$$

By continuity of the Ito integral this converges to $$0$$ as the mesh $$\to 0$$.

Edit:

The convergence of $$\sum_{i}(X_i-X_{i-1})^2$$ has to be understood as convergence in probability. See Remark I.V.$$1.19$$ from Revuz & Yor.

• First : thank you, is well seen (I didn't observe that and it helps). But I still have the problem that I don't see in which sense $\sum_{i}(X_i-X_{i-1})^2$ converges since $\sum_{i}(M_i-M_{i-1})^2$ doesn't converges a.s. – Todd Apr 27 at 16:34
• Indeed let me check but this is a well known result. – RScrlli Apr 27 at 16:38
• If I am not wrong convergence should be in probability. – RScrlli Apr 27 at 17:37
• then it works. Thank you :) Bu in my lecture, is written that $\sum_{i}(A_i-A_{i-1})^2$ converges in $L^2$ to $0$. You agree that it's wrong a priori, right ? – Todd Apr 27 at 17:42
• Hey @Todd see in my edit that I've found a source where this is explained! – RScrlli Apr 27 at 17:51