Note to the moderators: This question has been solved, and is indeed a valid question. Please post a comment explaning what needs to be elaborated on. Thank you very much! - random0620
Proposition 6.7 Let $X, Y \in \mathcal{P}, M \in \mathcal{M}_{\mathrm{c}}^2$ and $S \leq T$ be stopping times. Then $$ \begin{aligned} & \mathbb{E}\left[\left((X \cdot M)_{T \wedge t}-(X \cdot M)_{S \wedge t}\right)\left((Y \cdot M)_{T \wedge t}-(Y \cdot M)_{S \wedge t}\right) \mid \mathcal{F}_S\right]= \mathbb{E}\left[\int_{S \wedge t}^{T \wedge t} X_u Y_u \mathrm{~d}\langle M\rangle_u \mid \mathcal{F}_S\right] \end{aligned} $$
I am wondering how to solve this stochastic calculus question. In the question, $\mathcal{P}$ is the set of pre-visible processes. $\mathcal{M}_c^2$ is the set of square integrable martingales.
My attempt: I used the following formula :
$$(X \cdot M)_{min(T,t)} - (X \cdot M)_{min(S,t)} = \int_{min(S,t)}^{min(T,t)} X_s \, dM_s$$
$$ \mathbb{E}\left(\left[(X \cdot M)_{min(T,t)} - (X \cdot M)_{min(S,t)}\right]\left[(Y \cdot M)_{T^t} - (Y \cdot M)_{S^t}\right]\right) = \mathbb{E}\left[\int_{min(S,t)}^{min(T,t)} X_s \, dM_s \times \int_{min(S,t)}^{min(T,t)} Y_s \, dM_s\right]$$
Is there some way I can use the Ito formula to proceed in this question?
By stochastic integration by parts, which is a simple corollary of Ito's formula,
$$ X_t Y_t-X_0 Y_0=\int_0^t X_S d Y_s+\int_0^t Y_S d X_S+\int_0^t d X d Y $$
we have
$$\left((X \cdot M)_{T \wedge t}-(X \cdot M)_{S \wedge t}\right)\left((Y \cdot M)_{T \wedge t}-(Y \cdot M)_{S \wedge t}\right) \\=(X\cdot M)_{T\land t}(Y\cdot M)_{T\land t} -(X\cdot M)_{S\land t}(Y\cdot M)_{S\land t}\\ = \int_{S\land t}^{T\land t}X_sd(Y\cdot M)_{s} + \int_{S\land t}^{T\land t}Y_sd(X\cdot M)_{s} + \int_{S\land t}^{T\land t}d(X\cdot M)d(Y\cdot M) \\ $$
Now we have that
$$d(X\cdot M)d(Y\cdot M)= (XdM)(YdM) = XYdMdM = XYd\langle M\rangle.$$
Thus, $$ = \int_{S\land t}^{T\land t}X_sd(Y\cdot M)_{s} + \int_{S\land t}^{T\land t}Y_sd(X\cdot M)_{s} + \int_{S\land t}^{T\land t}X_sY_sd\langle M\rangle_s. \\$$
But after taking expectation, and applying Optional stopping which leaves only the variation term, we are left with
$$\mathbb{E}\left[(X\cdot M)|_{T\land t}(Y\cdot M)|_{T\land t} -(X\cdot M)|_{S\land t}(Y\cdot M)|_{S\land t}|\mathcal{F}_S\right]= \mathbb{E}\left[\int_{S\land t}^{T\land t}X_sY_sd\langle M \rangle_s|\mathcal{F}_S\right]$$