In the definition of a local martingale I have that for a filtered probasbility space $(\Omega, \mathcal{F},P,\mathbb{F})$. A local martingale is an adapted process M, such that there exists a sequence of increasing stopping times such that $\tau_{n}\rightarrow \infty$ a.s., and $M(t\wedge\tau_n)$ is a martingale for each n. This is also the definition given on Wikipedia.

I have a problem showing that if M is a local martingale, and $\tau$ is a stopping time, then $G(t)=M(t\wedge \tau)$ is a local martingale. The proof I have been given seems wrong:

By doobs optional sampling theorem we can show that for each n:

$E[G(t\wedge \tau_n)|\mathcal{F}_{s\wedge \tau}]=E[M(t\wedge \tau\wedge \tau_n)|\mathcal{F}_{s\wedge \tau}]=E[M((t\wedge \tau)\wedge \tau_n)|\mathcal{F}_{s\wedge \tau}]=_\text{optional sampling}M((s\wedge \tau)\wedge \tau_n)=G(s\wedge \tau_n)$.

In the theorem I used the two bounded stopping times $t\wedge \tau$ , and $s \wedge \tau$.

But this only shows that $E[G(t\wedge \tau_n)|\mathcal{F}_{s\wedge \tau}]=G(s\wedge \tau_n)$. But I need that $E[G(t\wedge \tau_n)|\mathcal{F}_{s}]=G(s\wedge \tau_n)$.

Do you see how to finish the proof? I am stuck.

  • 1
    $\begingroup$ Writing $M^{\tau_n}$ for the martingale $t\mapsto M(t\wedge \tau_n)$, observe that $G(t\wedge \tau_n)=M^{\tau_n}(t\wedge\tau)$, which is therefore a martingale because (thanks to Doob) a stopped martingale is a martingale. $\endgroup$ Commented May 14, 2016 at 23:59
  • $\begingroup$ @JohnDawkins Thank you, but how do you prove that a stopped martingale also is a martingale? I tried proving it, but I get the same problem as above. $\endgroup$
    – user119615
    Commented May 15, 2016 at 0:23
  • $\begingroup$ @JohnDawkins How do you prove that a stopped martingale is a martingale, unless it is almost surely right-continuous (and the filtration is right-continuous)? See math.stackexchange.com/questions/2329717/…. $\endgroup$
    – 0xbadf00d
    Commented Jun 20, 2017 at 13:33

1 Answer 1


In answer to the question in your comment on my comment: Let $X$ be a martingale, $T$ a stopping time, and $Y_t=X_{t\wedge T}$ the stopped process. By Doob, for $0\le s<t$, $E[Y_t\,|\,\mathcal F_{s\wedge T}]=Y_s$. Now let $A$ be an event in $\mathcal F_s$. On $A\cap \{T\le s\}$ we clearly have $Y_t=Y_s$. And $A\cap\{s<T\}$ is an element of $\mathcal F_{s\wedge T}$. Thus $$ E[Y_t\cdot 1_{A\cap\{T\le s\}}]=E[Y_s\cdot 1_{A\cap\{T\le s\}}]] $$ and (using optional sampling) $$ \eqalign{ E[Y_t\cdot 1_{A\cap\{s<T\}}] &=E[E[Y_t\,|\,\mathcal F_{s\wedge T\}}] 1_{A\cap\{s<T\}}]]\cr &=E[Y_s\cdot 1_{A\cap\{s<T\}}]\cr. } $$ Adding these two we obtain $$ E[Y_t\cdot 1_A]=E[Y_s\cdot 1_A], $$ which proves that $E[Y_t\,|\,\mathcal F_s]=Y_s$.

  • $\begingroup$ Thank you very much for your help, it was very good. $\endgroup$
    – user119615
    Commented May 15, 2016 at 22:22

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