Let $M$ be a $(\mathcal F_{t}^{B})_{t\geq 0}-$ local martingale starting in $0$ where $(\mathcal F_{t}^{B})_{t\geq 0}$ denotes the completed filtration of the Brownian motion. Now consider the reducing sequence $(\tau_{n})_{n}$ so that $M^{\tau_{n}}$ is a martingale. It is then stated that I may assume that $(M_{\tau_{n} \land t})_{t\geq 0}$ is a closed martingale (considering either $\tau_{n}$ or $\tau_{n}\land n$), $n \in \mathbb N$. I am not sure why closedness of the martingale $M^{\tau_{n}}$ or $M^{\tau_{n}\land n}$ holds. Could anyone help me to explain why this is the case?

Martingale $(M_{t})_{t\geq 0}$ is closed iff $M_{\infty}\in \mathcal{F}_{\infty}$ and integrable exists s.t. $E[M_{\infty}\lvert \mathcal{F}_{t}]=M_{t}$

  • $\begingroup$ For what purpose are you making that assumption? Is there a bigger lemma of theorem that's being proved with the help of the assumption? Also, in which book is this argument being made / paragraph found? $\endgroup$ Mar 7, 2021 at 18:11
  • $\begingroup$ Hint: If $ N=\{N_t,t\ge 0\} $ is a (right continuous) martingale, then $ N^K=\{N_{t\wedge K},t\ge 0\} $ is a right closed martingale for each $ K>0 $, since $ N_{t\wedge K}=\mathsf{E}[N_K|\mathscr{F}_t] $. $\endgroup$
    – JGWang
    Mar 8, 2021 at 11:53

1 Answer 1


First, we have that $\tau_n \wedge n$ is a bounded stopping time so $N_t := M_{(\tau_n \wedge n) \wedge t}$ is a closed martingale because $N_t = \mathbb{E}[N_n | \mathcal F_t]$ for all $t$, i.e. $N_\infty = N_n = M_{\tau_n \wedge n}$.

The reason that we may assume $(M_{\tau_n \wedge t})$ is closed is because if it isn't, we can replace $\tau_n$ with $\tilde{\tau}_n := \tau_n \wedge n$. Then $\tilde{\tau}_n$ is still a localizing sequence, and we showed above that $(M_{\tilde{\tau}_n \wedge t})$ is a closed martingale. Since we can always replace a localizing sequence with one that makes $(M_{\tau_n \wedge t})$ closed, we might as well just assume that the original localizing sequence does too.


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