before asking my question regarding of the martingale property of the stochastic integral I want to define some of the notations in the book I'm following:
- $\mathcal{M}^2$ is the set of continuous martingales $(M_t, t\in \mathbb{R}^{+} )$ in $L^2$ such that $M_0=0$.
- For $M \in \mathcal{M}^2$, let $L^2(M)$ denote the set of progressively measurable processes $(H_s, s \in \mathbb{R}^{+} )$ such that: $$ \mathbb{E} \left[ \int_{0}^{T} H_s^2 d\langle M \rangle_t \right] < \infty $$
I would like to prove that, if $M, N \in \mathcal{M}^2$, $H \in L^2(M) $ and $K \in L^2(N)$, then for every finite stopping time $T$, we have:
$$ \mathbb{E} \left[ \left( \int_0^{T} H_s dM_s \right) \left( \int_0^{T} K_s dN_s \right)\right] = \mathbb{E} \left[ \int_0^{T} H_sK_s d\langle M,N \rangle_s \right]$$
My first approach was to prove that $L_t := (H \cdot M)_{t}(K \cdot N)_{t} - \langle H\cdot M,K \cdot N \rangle_t$ is a martingale $L^2$ (ie $L_t \in \mathcal{M}^2$) and so we can apply the stopping theorem and knowing that $L_0=0$ we would have proven the equality. It is not very clear that it is a martingale, but by definition of the quadratic variation, it is a local martingale. I don't know how to prove that $\mathbb{E} [L_t^2] < \infty$. I've tried with Kunita-Watanabe inequality but I don't find the right conditions that $L_t$ should verify for being able to apply the stopping theorem.
On the other hand, I've been thinking that given the stopping time $T$ is finite, then if $L_t$ is a martingale, the stopped martingale $L_t^T$ is bounded by the value of the martingale in $T$ so, $L_t^T \in \mathcal{M}^2$ and we would have the desired equality.
I'm not sure if my reasoning make sense, and if it possible to prove my first idea. Every help is much appreciated. Thank you!
I do not believe this result holds, consider for example two independent one dimensional Brownian motions $B_s$ and $B'_s$. Their quadratic covariation is equal to zero at every timepoint, whilst $⟨B'⟩_s = ⟨B ⟩_s = s$, meaning the right side of your equation is always zero while the left side isn't necessarily.