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Proposition: Let $N_t$ be an $\mathcal{F}$-Poisson process and $M_t=N_t-\lambda t$ its compensated process. Then for any $\mathcal{F}$-predictable bounded process $H_t$, the stochastic integral $$(H\star M)_t:=\int_0^t H_{s}dM_s=\int_0^t H_{s}dN_s-\lambda\int_0^t H_{s}ds\\ $$ is a martingale.

Here $N_t$ is a counting process defined as $$ N_t=\sum_{n\ge 1}\mathbb{1}_{\{T_n\le t\}}=\sum_{n\ge 0}n\mathbb{1}_{\{T_n\le t<T_{n+1}\}}$$ where $(T_n,n\ge0)$ is a sequence of random times at which the jumps of $N_t$ happen. Moreover $\mathbb{P}(N_t=n)=e^{\lambda t}\frac{(\lambda t)^n}{n!}$.

The martingale $M_t$ is the compensated process associated to $N$, defined as $M_t=N_t-\lambda t$.

This proposition is interesting because it extends the martingale property of $M_t$ to the stochastic integration with respect to $M_t$ for bounded predictable integrands $H_t$.

I understand that this property can be proven by considering first simple integrands $H_t$ of the form $H_t=K_S \mathbb{1}_{]S,T]}(t)$, where S and T are two stopping times and $K_S$ is $\mathcal{F}$-measurable. Then, one can pass to the limit for general $H_t$.

But could anyone give a complete proof (or a reference to a complete proof) of these statements? This would also be very instructive as it is a very good example of working with Poisson Processes, Martingales, Stopping Times, Convergence Theorems and Stochastic Integration.

Moreover what would happen in the case $H_t$ is adapted, instead of predictable? Because this proposition seems to imply e.g. that the process $\int_0^t N_{s-}dM_s$ is a martingale, but the the process $\int_0^t N_{s}dM_s$ is not a martingale.

Pasriv
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2 Answers2

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This might be helpful: Theorem 29 on page 173 of Protter's "Stochastic integration and differential equations":

Let $M$ be a local martingale and let $H\in \mathcal P$ be locally bounded. then the stochastic integral $H\cdot M$ is a local martingale.

I'm reading that book, but haven't got that far yet, will be interested to see if someone can provide a simple proof just for this special case in this post.

Jay.H
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  • Yes, I guess the question I asked is actually a refinement of this property, since $M$ in this case is a martingale. I will check the proof you're citing to see if it can be adapted to this particular case. However there may be more to it, because the predictability of the integrand function is important here, since this property does not hold generally for adapted processes $H_t$. – Pasriv May 23 '16 at 19:23
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    I should have mentioned, $\mathcal P$ stands for predictable process. Seems to me, that theorem almost addresses your question, the only gap is to show the integral is a martingale, given that we already know it is a local martingale by the theorem. But that should be simple, given that your $H$ is bounded. – Jay.H May 23 '16 at 19:31
  • Thanks for pointing me to Protter's book. Actually there are also two interesting related examples concerning this property, even if there is not a proof of this precise statement. At page 45 there is a good example of this, where the theorem is proven when $H_t$ is bounded with continuous sample paths. At page 77 there is an example that show that for general bounded adapted càdlàg $H_t$, the property does not hold (like I was writing in the last note of the post). I am also interested in a proof for the simpler example in this post, as it could also help to understand the general case. – Pasriv May 23 '16 at 20:45
  • For the proof of your proposition, I think it is not difficult if we are allowed to use Theorem 29. For counter example in general case, you may check out the Emery's example on page 178 ( I haven't look it through carefully yet). – Jay.H May 23 '16 at 21:26
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Proof

First, the compensated process of Poisson prcocess is a strict martingale. To see this, note that

$$\begin{aligned}E(N_t|\mathcal{F}_s)&=E[N_t-N_s+N_s|\mathcal{F}_s]\\&=E[N_t-N_s]+N_s\\&=\lambda(t-s) + N_s\end{aligned}$$

which means

$$E[N_t-\lambda t|\mathcal{F}_s]=N_s-\lambda s$$

Then, just apply the fact that a stochastic integral w.r.t. a martinagle is still a martingale. This is a famous result that you can find in all books on stochastic integration. For example, Thm. 6.5.8 of Introduction to Stochastic Integration by Hui-Hsiung Kuo. Your $H$ is predictable and bounded, hence clearly in $L^2_{pred}(M)$.

About the second question

I don't think so, as the stochastic integration is only well-defined for $H$ being predictable.

Mingzhou Liu
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  • There are non predictable integrands that can be perfectly integrated w.r.t. a compensated Poisson process. What OP wrote in the last sentence is true when $H$ is a Poisson process continuous from the right. – Kurt G. Apr 19 '24 at 12:36