Revuz and Yor - Continuous Martingales and Brownian Motion 1999

Chapter IV Proposition 1.13

it is proven, that for a continuous local martingale $M_t$ the intervals of constancy are equal with those of the predictable quadratic variation $<M>_t$ or optional quadratic variation $[M]_{t}$ (since they coincide due to the continuity of the local martingale).

I wonder if this stays true for $M_{t}$ being just càdlàg. I guess no.

So lets consider this setup: Given a square integrable Martingale $X_t=F_t-a\cdot K_{t}$ with predictable quadratic variation $b\cdot K_{t}$ where $a,b$ are constants and $K_{t}$ is continuous but $F_{t}$ only càdlàg. With the aim to conclude from $K_{t}$ being constant on some interval (predictable quadratic variation is continuous process of $K_{t}$ being constant) implies that $X_{t}$ is constant on that interval and thus $F_{t}$ on the interval. Where $K_{0}=0$, $K_{t}\rightarrow \infty$ a.s. and a non decreasing process.

  • $\begingroup$ did you try if it works with a Poisson process minus its compensator $\endgroup$
    – MJ73550
    Apr 28, 2016 at 12:21
  • $\begingroup$ @MJ73550 No, but can you explain your idea behind this? Edit: I edited from where i got the problem, maybe it is more clear then. $\endgroup$
    – ziT
    Apr 28, 2016 at 13:08
  • $\begingroup$ let $N_t$ be a Poisson process of intensity $\lambda$, $M_t=N_t-\lambda t$ is a càdlàg martingale, $<M>_t=\lambda t$ (you compute $\mathbb{E}(M^2_t-\lambda t|\mathcal{F}_s)$). Now $[M]_t=\sum_{s\leq t}(\Delta N)^2_s=\sum_{s\leq t}\Delta N_s = N_t$. So it is different, and have no intervals of constancy in common $\endgroup$
    – MJ73550
    Apr 28, 2016 at 14:02

1 Answer 1


It cannot work for càdlàg martingales.

Let $N_t$ be a Poisson process of intensity $\lambda$,

set $M_t=N_t-\lambda t$. It is a càdlàg martingale,

$<M>_t=\lambda t$ (you compute $\mathbb{E}(M^2_t-\lambda t|\mathcal{F}_s)$).

But going back to definition of $[M]$, you get $[M]_t=\sum_{s\leq t}(\Delta N)^2_s=\sum_{s\leq t}\Delta N_s = N_t$.

  • $\begingroup$ Thanks so far. This is a counterexample, which showed me, that i have missed the requirements of $K_{t}$ being increasing, but not strictly increasing. Can you look at the cited book p.48. I really don't get it. Since then the whole theory in the chapter relies on a mistake. $\endgroup$
    – ziT
    Apr 28, 2016 at 14:22
  • $\begingroup$ I do not know this book neither his author Kuchler, but I do know Revuz and Yor. $\endgroup$
    – MJ73550
    Apr 28, 2016 at 14:51
  • $\begingroup$ @MJ73550 : Although in this case M and <M> have the same intervals of constancy i.e. none. Best regards $\endgroup$
    – The Bridge
    Apr 29, 2016 at 6:45

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