Definition A cadlag $L^2$-martingale $M$ is called purely discontinuous, if $M \cdot N$ is a martingale for any continuous $L^2$-martingale $N$. (If not otherwise mentioned, I always assume that the martingales start in $0$, i.e. $M_0 = N_0 =0$).

I would like to prove the following statement:

Let $M$ a purely discontinuous martingale. Then the quadratic variation of $M$ is given by $$[M]_t = \sum_{s \leq t} (\Delta M_s)^2$$ where $$[M]_t := \text{UCP}-\lim_{|\Pi| \to 0} \sum_{t_j \in \Pi} (M_{t_j}-M_{t_{j-1}})^2$$ for partitions $\Pi = \{t_0<\ldots<t_n=t\}$ of $[0,t]$.

Using the formula $$[M]_t = M_t^2 - 2 \int_0^t M_{s-} \, dM_s$$ it's not difficult to see that $$\Delta [M]_t = (\Delta M_t)^2$$ Moreover, since $[M]$ is of bounded variation, we have $$[M]_t = [M]_t^c + \sum_{s \leq t} \Delta [M]_s = [M]_t^c + \sum_{s \leq t} (\Delta M_s)^2$$ where $[M]_t^c$ denotes the continuous part of $[M]$.

Consequently, I would like to show that $[M]_t^c = 0$, but I don't see how to proceed. Since $M$ is purely discontinuous, we get $$[M,N]_t = 0$$ for any continuous $L^2$-martingale $N$, but I don't have a clue how this should help me.


(Notation: $\Delta M_s := M_s - M_{s-}$; ucp: uniform in probability)

  • $\begingroup$ I guess you are right and as you found a reference I erased my comments. Regards. $\endgroup$
    – TheBridge
    Commented Aug 29, 2013 at 12:30
  • $\begingroup$ @TheBridge Thanks for your effort, nevertheless! $\endgroup$
    – saz
    Commented Aug 29, 2013 at 13:16

1 Answer 1


In the meantime, I found a proof in the following book

Jacod/Shiryaev: Limit theorems for stochastic processes; Lemma I.4.51

and apparently the result is rather deep.

I'm still interested in alternative proofs.


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