$M_t$ is a martingale with respect to $\mathcal{F _t}$ for $t \geq 0$ and $Z$ is a bounded $\mathcal{F_r}$ measurable random variable. $0\leq r < s <\infty$. I want to show that $Z( M_{s\wedge t}-M_{r\wedge t})$ is a martingale.

I started by introducing an additional time variable p such that $r \leq u <s$ and then I was thinking of looking at $E[Z( M_{s\wedge t}-M_{r\wedge t})|F_{u}]$ but I am unsure to deal with product. Does anyone have any advise?


Do not forget to mention the fact that $Z( M_{s\wedge t}-M_{r\wedge t})$ is measurable and integrable in the sigma-algebra $F_s$.

Otherwise this is a good start: now as $Z$ is $F_u$ measurable, you have

$$E[Z( M_{s\wedge t}-M_{r\wedge t})|F_{u}] = Z E[ M_{s\wedge t}-M_{r\wedge t}|F_{u}] = Z E[ M_{s\wedge t}|F_{u}] -M_{r\wedge t} $$and as the stopped martingale is a martingale as well, $$ E[ M_{s\wedge t}|F_{u}] = M_{u\wedge t} $$

If you put that together, you have what you want.

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