# Continuous local martingale of finite variation is constant

Is a continuous local martingale $M$ of finite variation constant?

We know that there exists a sequence of stopping times $T_n\nearrow \infty$ a.s. as $n\to\infty$ such that the stopped process $M^{T_n}$ is a continuous bounded martingale. And $M^{T_n}$ has finite variation, because $M$ has finite variation.

Since if $X$ is a continuous bounded martingale with finite variation, then $X=X_0$ a.s. So $M_t^{T_n}=M_0^{T_n}$ a.s. $\forall t\ge 0$ and by letting $n\to\infty$ we obtain that $M\equiv M_0$.

1. Is my reasoning correct?

2. The statement "continuous bounded martingale with finite variation implies constant" is a lemma from my notes, but we did not prove it. How would one prove it?

• For 2: If a martingale is continuous and bounded, then it's also a continuous local martingale, no? Jan 24, 2014 at 10:45
• One way is to show, that finite variation implies quadratic variation equals zero and then use this proof: math.stackexchange.com/q/473786/445105 Feb 26, 2020 at 22:19

Your intuition is good, but there are some technicalities you should be careful about. WLOG $M_0 = 0$. Start by assuming $M$ is a continuous martingale of bounded variation. Then if $B$ is a bound on the variation of $M$, and $(t_i)$ is a partition of $[0,t]$,

$$E[M_t^2] = E\sum (M_{t_{i+1}}-M_{t_i})^2 \leq B E[\sup_i|M_{t_{i+1}}-M_{t_i}|].$$

Since $M$ is continuous the supremeum tends to $0$ as the partition size goes to $0$. Moreover, the supremum is dominated by $B$, and hence dominated convergence tells us that $E[M_t^2] = 0$, in particular $M_t=0$ a.s. Let $t$ run over rationals and use the continuity of $M$ to conclude that $M=0$.

Next, if $M$ is a continuous local martingale, take a localizing sequence of stopping times $(\tau_n)$. We then find that $M^{\tau_n}_t = 0$ for all $n$, and so taking the limit in $n$ gives $M_t = 0$ a.s. Again let $t$ range over rationals.

• @Serena, check that $M$ having bounded variation implies $M_t \in L^2$. For a martingale with each $M_t \in L^2$ one has $E[M_t^2] = E[(\sum M_{t_{i+1}}-M_{t_{i}})^2]$. Expand and check that the martingale property implies that all cross terms have expectation zero. The $B$ bound is from $\sum |M_{t_{i+1}}-M_{t_i}| \leq B$ by definition of $B$ being a bound on the variation of $M$. The sup bound is saying any individual $|M_{t_{i+1}}-M_{t_i}|$ is at most the sup over all such terms. Nov 15, 2017 at 20:50
• No, the variation of a function $f$ over an interval $[a,b]$ is defined to be the supremum over all finite partitions of $[a,b]$ of $\sum |f(t_{i+1}) - f({t_i})|$. See en.wikipedia.org/wiki/Bounded_variation Nov 17, 2017 at 4:42
• How do you get bounded variation and thus the bound B if you only have finite variation? Feb 26, 2020 at 16:04
• @roi_saumon We have $E[M_t^2] =0$ which implies $M_t = 0$ a.s. But the measure zero set on which $M_t = 0$ depends on $t$. Since the rationals are countable, we still have $M_q = 0$ for all rational $q$, a.s., whereas we could not say this if we let $t$ vary over the reals. Since the rationals are dense in the reals, the continuity of $M$ then allows us to conclude $M=0$ a.s. Any countable dense set would have worked, not just $\mathbb{Q}$. Aug 30, 2021 at 16:32
• @roi_saumon On the event that $\tau_n \to \infty$, one has that $\min(\tau_n, t) \to t$ and hence $M_t = \lim_n M_t^{\tau_n} = \lim_n 0 = 0$, so $M_t = 0$ a.s. Aug 30, 2021 at 23:55

I am not sure if it is that simple. I detest posting a link only answer but I feel I am not adding any value by posting anything in addition to this excellent link.

see theorem 3 and lemma 4 of

http://almostsure.wordpress.com/2010/04/01/continuous-local-martingales/