# Quadratic variation and stopping time.

Let $X_t$ be a strict local martingale and let $S$ and $T$ be stopping times with $S \leq T$. Prove that $[X]_S=[X]_T$ implies that $X_t$ is almost surely constant on $[S, T]$, where $[X]_S$ and $[X]_T$ is the quadratic variation of the stopped process.

-

For continuous local martingales $X_t$, it is known (theorem 3) that $X_t$ and $[X]_t$ have the same intervals of constancy almost surely, by conditioning over the event $[S,T]$ you get the result in that case.

Nevertheless and contrary to what I said before editing the answer, I have a found simple counterexample for general local martingale (i.e. with jumps).

Let's take the (local) martingale $X_t= N_t-t$, (where $N_t$ is a Poisson process of intensity 1), and for $S=S_1$ the first jump time of $X$ and $T=S+1$ (here $S_i$ is the $i$-th jump time of $N_t$), then the event $A=\{\omega\in\Omega s.t. T<S_2\}$ has strictly positive probability equal to $e^{-1}$ (it is widely know that inter-arrival times for Poisson process of intensity $\lambda$ follows an exponential law of parameter $\lambda$).

So on $A$, we have $[X]_S=1=[X]_T$, but $X_S=1-S$ and for $t\in[S,T]$ we have $X_t-X_S=(1-t)-(1-S)=S-t$ which is not constant but continuous of finite variation, and so $X_t$ isn't constant on the interval $[S,T]$ almost surely as $A$ has positive probability.

This shows that the conclusion is wrong in general case.

So I think, that you omited to mention that your problem was only in a continuous process setting, as it is the only place where it is true.

Edit : I felt that my preceding wrong proof was fishy sorry for that.

Edit 2: Thanks to Didier Piau for pointing out a clarification of the argument.

Best Regards

NB: Notice that the result (in continuous process setting) is more general as it works for all continuous local martingales and not only for continuous strict local martingales.

-
Nice post.   –  Did Dec 15 '11 at 13:39
@Didier : Thank's –  TheBridge Dec 15 '11 at 20:59
Yes, nice answer. However, the question itself is not completely clear to me. As you show, ($[X]_S=[X]_T$ ⇒ $X_S=X_T$) (almost surely) is false in general. However, ($[X]_S=[X]_T$ (a.s.)) ⇒ ($X_S=X_T$ (a.s.)) is true. The answer depends on where you put your "almost surely"s. –  George Lowther Dec 15 '11 at 23:30
@ George Lowther : I couldn't have answered this question if I hadn't read your blog so Thank's to you too ;-). –  TheBridge Dec 16 '11 at 8:33