Let $\tau$ be a stopping time and $(M_n)_{n \in \mathbb{N}_0}$ be a martingale with $\mathbb{E}(M_n^2)<\infty$ for any $n \in \mathbb{N}_0$. Show that, if $\langle M \rangle_{\tau} = 0$ (where it means the quadratic variation) a.s. then $$\mathbb{P}(M_{\tau \wedge n} = M_0 \, \, \text{for any $n \in \mathbb{N}_0$})=1.$$

I cannot understand the point, in fact $$\mathbb{P}(M_{\tau \wedge n} = M_0 \, \, \text{for any $n \in \mathbb{N}_0$})=1$$ I think it means that the stopping time is $\tau=0$ but it makes no sense. Could someone help me to resolve this exercise?

  • $\begingroup$ It means that the martingale is constant (for fixed $\omega \in \Omega$) up to time $\tau$, i.e. $M_0(\omega) = M_1(\omega) = \ldots = M_n(\omega)$ for all $n \leq \tau(\omega)$. $\endgroup$
    – saz
    Jan 23, 2019 at 18:57
  • $\begingroup$ Perfect I have understood. Could you help me also with the assumption? Because I know that the quadratic variation equal to 0 means that also Mn^2 is a martingale, but what change if there is τ instead of n? $\endgroup$ Jan 23, 2019 at 19:22

1 Answer 1



  1. Use the optional stopping theorem to show that $$X_n := M_{n \wedge \tau}^2 - \langle M \rangle_{n \wedge \tau}$$ is a martingale; in particular $$\mathbb{E}(M_{n \wedge \tau}^2) - \mathbb{E}(M_0^2) = \mathbb{E}(\langle M \rangle_{n \wedge \tau}). \tag{1}$$
  2. Deduce from $\langle M \rangle_{\tau} = 0$ that $\langle M \rangle_{n \wedge \tau} = 0$ for all $n \in \mathbb{N}_0$.
  3. Show that $$\mathbb{E}((M_{n \wedge \tau}-M_0)^2) = \mathbb{E}(M_{n \wedge \tau}^2)-\mathbb{E}(M_0^2), \qquad n \in \mathbb{N}_0.$$ (Expand the square and then use the tower property of conditional expectation + the martingale property.)
  4. Combining the first three steps shows that $$\mathbb{E}((M_{n \wedge \tau}-M_0)^2)=0$$ for all $n \in \mathbb{N}_0$. Conclude.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .