f I$M,N$ are local martingale, then so is $N+M$ but not $NM$.

Question

Let $M$ and $N$ being local martingales.

• I want to prove that $M+N$ is a local martingale. Let $(\tau_n)$ and $(\sigma _n)$ sequences s.t. $\sigma _n,\tau_n\nearrow \infty $ and s.t. $(M_{t\wedge \sigma _n})$ and $(N_{t\wedge \tau_n})$ are martingales for all $n$. Which sequence of increasing stopping time $(\kappa_n)$ we'll make $(M_{t\wedge \kappa_n}+N_{t\wedge \kappa_n})$ a martingale ? Since a priori $(M_{t\wedge \tau_n})$ or $(N_{t\wedge \sigma _n})$ are a priori not martingale...

• Can someone provide local martingales $M$ and $N$ s.t. $MN$ won't be a local martingale ?

Answer 1

Hint

For the sum, notice that if $(M_t)$ is a right-continuous martingale, then so is $(M_{t\wedge \tau})$ for any stopping time. Therefore, $\kappa_n=\sigma _n\wedge \tau_n$ indeed make the job.