Suppose that $Y$ is a pure jump process with $N_t$ jumps in $(0,t]$ and $E[N_t]<\infty$. Denote the jump times by $T_i$. Let $W$ be a Brownian motion. If $T_0=0$, then \begin{equation} M_t=\int_0^t Y_s\,dW_s=\sum_{i=0}^{N_t-1} Y_{T_i}(W_{T_{i+1}}-W_{T_i})+Y_t(W_t-W_{T_{N_t}}). \end{equation} The process $M$ is continuous. I want to show that $M$ is a local martingale by using the sequence $\tau_n=\inf\{t: |M_t|>n\}$. That is, I have to show that $\{M_{\tau_n\wedge t}\}$ is a martingale. I have read that one can use the above sequence of stopping times for continuous local martingales. But why or how?


1 Answer 1


I guess you could use the sequence $\tau_n=\inf\{t\,:\, N_t=n\}$ and then Corollary 6.14 (or 7.14 depending on the edition, it is called "martingale transforms") from Kallenberg's Foundations of Modern Probability yields the result.

The corollary states that $Y_t(W_t-W_{T_{N_i}})$ and $$\sum_{i=0}^{n}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$ are martingales for all (fixed / deterministic) $n\in \mathbb N$.

But in your case, you need that $$\sum_{i=0}^{N_{t-1}}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$ is a local martingale for random $N_{t-1}$. This is why you need to stop the process. With the stopping time suggested above, you obtain $$M_t^{T^n}=\sum_{i=0}^{n-1}Y_{T_i}(W^{T_{i+1}}_t-W^{T_i}_t)$$ which is a martingale.

Of course, you could also use the well known fact, that a stochastic integral with respect to continuous local martingale is always a local martingale. Then you still need to show, that stopping at $\tau_n=\inf\{t\,;\,|M_t|>n\}$ yields a uniformly integrable martingale. Because of the continuity of the process $M_t$, you get that $M_t^{\tau_n}$ is bounded by $n$ and any bounded local martingale is a uniformaly integrable martingale. Hence the result follows.

  • $\begingroup$ This would imply that $M$ is even a martingale, not just a local martingale, which makes me sceptical about this answer. $\endgroup$
    – Sofia
    Nov 26, 2018 at 18:05
  • $\begingroup$ You need to stop the process $M$ in order to make the sum consist of a fix number of summands. Without stopping the number of summands is random and thus you can't apply the corollary from Kallenberg. Thus you have only shown that $M$ is a local martingale. I don't know whether $M$ might even be a martingale. I can imagine that you need more information about the distribution of $N$ to determine whether it is only a local martingale or even a true martingale. $\endgroup$
    – Agnetha
    Nov 26, 2018 at 18:45
  • $\begingroup$ Then I didn't understand how you would use the corollary. Could you please explain that a bit more precisely? $\endgroup$
    – Sofia
    Nov 27, 2018 at 9:37
  • $\begingroup$ I edited my answer above. $\endgroup$
    – Agnetha
    Nov 27, 2018 at 10:08

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