I am looking for a direct proof of the fact that, roughly speaking, if $S=S_0+A+M$ is an $L^2$ semimartingale, and $M$ (the martingale part) has the martingale representation property, then for any square integrable random variable in $\mathcal{F}_T$, $X=X_0 + \int_0 ^T K_t d S_t$ for some predictable process $K$ (this would be true via the beloved "complete market"/unique martingale measure scenario from financial math, but I am looking for a proof that avoids a change of measure). The filtration here could be the natural filtration of $S$, or maybe a bigger one. In any case, all the usual assumptions are supposed to hold.

In the process of trying, I ran into a problem I can't figure out (I am getting rusty). I'll post a simplified version, and, if there's interest, I'll go to the full-fledged version. Everything here is cadlag and square integrable.

Suppose we partition $[0,T]$ in $n$ equal subintervals, $\left[ t_j,t_{j+1} \right]$, $j=0, \ldots, n-1$ Suppose that, besides the semimartingale $S_0 + A + M$, we have a finite-variation predictable process $K$. I would like to show that $\int_0^T K_s dA_s - \sum_{j=1}^{n-1} E \left[ \int_{t_j}^{t_{j+1}} K_s d A_s \left| \mathcal{F}_{t_j} \right. \right]$ goes to zero (in $L^2$, say), as $n \rightarrow \infty$.

As I said, I am getting rusty, so I am not sure what additional assumptions would be necessary (a back of the envelope calculation convinced me that this should work for Brownian Motion with drift, but I'd hope it's much more general), as I suppose that something is missing here.

Thank you.


1 Answer 1



I think you should have a look at Protter's book "Stochastic Integration and Differential Equations" in the $L^2$ case I think that there is a direct proof there of this fact.


  • $\begingroup$ Thank you. I have been looking at Philip's book (and Liptser-Shirayev's) but didn't find what I was looking for. I guess I'll try to dig deeper. $\endgroup$
    – Federico
    Oct 15, 2010 at 17:16

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