# How can I show that the stochastic integral of a jump process w.r.t. Brownian motion is a local martingale by using this special localizing sequence?

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 $$$$M_t=\int_0^t Y_s\,dW_s=\sum_{i=0}^{N_t-1} Y_{T_i}(W_{T_{i+1}}-W_{T_i})+Y_t(W_t-W_{T_{N_t}}).$$$$ The process $$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?

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.
• This would imply that $M$ is even a martingale, not just a local martingale, which makes me sceptical about this answer. Nov 26, 2018 at 18:05
• 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. Nov 26, 2018 at 18:45