I am reading the Dubins-Schwarz theoem on Brownian motion and stochastic calculus. It says, given a continuous local martingale $M$ such that $\lim_{n\to\infty}[M]_t = \infty$ a.s., where $[M]_t$ denotes the quadratic variation of $M$. For each $s \geq 0$, the book defines a stopping time $$T(s)=\inf\{t \geq 0: [M]_t > s\}$$ and says the stopped process $M_{t\wedge T(s)}$ is a martingale.

I would like to ask how can we see the stopped process is indeed a martingale? Should we use the fact that a class $DL$ local martingale is a martingale? Many thanks


1 Answer 1


Let $(\sigma_n)_{n \in \mathbb{N}}$ be a localizing sequence of the local martingale $M$, i.e. an increasing sequence of stopping times such that $\sigma_n \to \infty$ and $(M_{t \wedge \sigma_n})_{t \geq 0}$ is a martingale for each $n \in \mathbb{N}$.

Applying the optional stopping theorem, we find that $(M_{t \wedge \sigma_n \wedge T(s)})_{t \geq 0}$ is a martingale, i.e.

$$\mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)} \mid \mathcal{F}_u) = M_{u \wedge \sigma_n \wedge T(s)} \tag{1}$$

for all $u \leq t$. Since $M$ has continuous sample paths, the right-hand side converges to $M_{u \wedge T(s)}$ as $n \to \infty$. If we can show that

$$\lim_{n \to \infty} \mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)} \mid \mathcal{F}_u) = \mathbb{E}(M_{t \wedge T(s)} \mid \mathcal{F}_u)\tag{2}$$

this proves that $(M_{t \wedge T(s)})_{t \geq 0}$ is a martingale. To prove $(2)$, we note that by Doob's maximal inequality

$$\begin{align*} \mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge \sigma_n \wedge T(s)}|^2 \right) &\leq 4 \mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)}^2) = 4 \mathbb{E} \big( [M]_{t \wedge \sigma_n \wedge T(s)} \big) \leq \mathbb{E}([M]_{T(s)}) \leq 4s. \end{align*}$$

By the monotone convergence theorem this implies

$$\mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge T(s)}|^2 \right) \leq 4s < \infty.$$


$$\mathbb{E} \left( \sup_{n \in \mathbb{N}} |M_{t \wedge \sigma_n \wedge T(s)}| \right) \leq \sqrt{\mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge T(s)}|^2 \right)} < \infty.$$

Now $(2)$ follows from the dominated convergence theorem.

  • $\begingroup$ Many thanks for your help. I have one more question: for $M_{u\wedge\sigma_n\wedge T(s)} \to M_{u\wedge T(s)}$, is the right-continuity of $M$ sufficient (and necessary) to show the convergence? $\endgroup$
    – Kenneth Ng
    Mar 15, 2017 at 6:13
  • $\begingroup$ @KennethNg Actually, we don't need continuity for this. Since $\sigma_n \to \infty$ we have $u \wedge \sigma_n \wedge T(s) = u \wedge T(s)$ for $n$ sufficiently large, and so $$M_{u \wedge \sigma_n \wedge T(s)} = M_{u \wedge T(s)}$$ for $n$ sufficiently large. (Mind that this "sufficiently large" depends on $\omega$, $s$ and $u$.) $\endgroup$
    – saz
    Mar 15, 2017 at 6:51
  • $\begingroup$ saz: In (1), shouldn't the conditioning be on $\mathcal{F}_{u \wedge T(s)}$? $\endgroup$
    – jpv
    Mar 15, 2017 at 11:50
  • $\begingroup$ @jpv Why do you think so? $\endgroup$
    – saz
    Mar 15, 2017 at 12:15
  • $\begingroup$ The OST applied on the martingale $M_{t \wedge \sigma_n}$ for the stopping times $t \wedge T(s)$ and $u \wedge T(s)$ results in that expression. $\endgroup$
    – jpv
    Mar 15, 2017 at 12:28

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