Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\ge0}$ be a right-continuous filtration of $\mathcal A$
- $\sigma,\tau,\rho$ be $\mathcal F$-stopping times with $\sigma\le\tau\le\rho$
- $\eta,\xi:\Omega\to\mathbb R$ be almost surely bounded $\mathcal F_\sigma$- and $\mathcal F_\tau$-measurable random variables on $(\Omega,\mathcal A,\operatorname P)$ respectively
- $M:\Omega\times[0,\infty)\to\mathbb R$ be an almost surely right-continuous $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $M_t\in L^2(\operatorname P)$ for all $t\ge0$
Let $t\ge0$. How can we show that $\operatorname E\left[\left|\eta(M_t^\tau-M_t^\sigma)+\zeta(M_t^\rho-M_t^\tau)\right|^2\right]=\operatorname E\left[\left|\eta(M_t^\tau-M_t^\sigma)\right|^2\right]+\operatorname E\left[\left|\zeta(M_t^\rho-M_t^\tau)\right|^2\right]$?
I could imagine that we can condition on $\mathcal F_\sigma$ and use $\sigma\le\tau$ together with the Optional Sampling theorem, but I'm not sure how exactly we're able to apply this approach.
This is equivalent to prove that $\mathbb{E} \left[ \eta \zeta (M_t^\tau - M_t^\sigma)(M_t^\rho - M_t^\tau)\right] = 0$. Since $\sigma \le \tau$ then $\mathcal{F}_\sigma \subset \mathcal{F}_\tau$, and $\mathbb{E} \left[ \eta \zeta (M_t^\tau - M_t^\sigma)(M_t^\rho - M_t^\tau)\Bigg | \mathcal{F}_\tau\right] = \eta \zeta (M_t^\tau - M_t^\sigma) \mathbb{E} \left[(M_t^\rho - M_t^\tau) \Bigg | \mathcal{F}_\tau\right]$. Using the Optional sampling theorem we have $\mathbb{E} \left[M_{\rho\wedge t}\Bigg | \mathcal{F}_\tau\right] = M_{\tau \wedge t}$. Can you conclude from here ?