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In the book Stochastic calculus of Baldi, there is the following lemma :

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For the last equality, they made a quite long and complicate proof. But, don't we simply have that $$\int_a^b X_sdW_s\quad \text{and}\quad \int_a^bX_s^2ds$$ are independent of $\mathcal F_a$ ? And thus, using Itô isometry $$\mathbb E\left[\left(\int_a^bX_sdW_s\right)^2\mid \mathcal F_a\right]=\mathbb E\left[\int_a^bX_s^2ds\right]=\mathbb E\left[\int_a^b X_s^2ds\mid \mathcal F_a\right]\ \ ?$$

$M^2(0,T)$ is the set of process s.t. $X_t$ is progressively measurable and s.t. $$\mathbb E\int_0^TX_s^2ds<\infty .$$

Walace
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    I may be wrong, but I don't think that $\int_a^b X_sdW_s$ and $\int_a^bX_s^2ds$ are independent of $\mathcal F_a$. – Todd Apr 27 '20 at 10:05
  • does "elementary process " means that it's a "simple" or "piecewise" process? – Chaos Apr 27 '20 at 10:16
  • Is $\mathcal F$ the natural filtration of your BM? Do you assume $X(a)$ is independent to $F_a$? if not, then you don't have independence! – Chaos Apr 27 '20 at 10:17
  • @RScrlli: yes elementary is simple process. But this result still hold for more general process. $\mathcal F$ is an admissible filtration (but we can take the natural filtration). and no $X_a$ is not independent of $\mathcal F_a$ (but if $X_a\in \mathcal F_a$ how can it be independent of $\mathcal F_a$ ?) – Walace Apr 27 '20 at 10:57
  • I am not saying that if $X_a$ is $F_a$ measurable then it's independent! If you assume that $X$ is adapted I don't think you have the independence you claimed on the question! – Chaos Apr 27 '20 at 11:14
  • I guess the fact that X is adapted is implicit in the definition of $M^2$ right? I overlooked that, though you should add the definitions because notation may differ from book to book! – Chaos Apr 27 '20 at 11:19
  • Thanks. I add the definition of $M^2$. – Walace Apr 27 '20 at 11:47

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