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I am trying to understand why semimartingales are the most general possible class of stochastic integrators. (I was hoping that this question would give me my answer, but it didn't.)

I thought at first it was because they were the most general class of processes with defined quadratic variation.

(Because I thought local martingales are supposed to be the most general class for which the quadratic variation was non-zero and could not be decomposed into the sum of non-trivial processes, one with zero quadratic variation, and because adapted processes of locally bounded variation have zero quadratic variation.)

However, it turns out that this is not the content of the Bichteler-Dellacherie theorem (as I had expected). Furthermore, there are apparently counter-examples involving fractional Brownian motion (which I do not understand at all) Quadratic variation - Semimartingale

The definition of the stochastic integral, when moving from Brownian motion for which the differential is just $dt$ because the quadratic variation is $t$, to arbitrary local martingales, and then semimartingales, clearly has something to do with quadratic variation, but I feel like I must be missing some part of the story if semimartingales are not the most general processes for which quadratic variation exists, is defined, and can be non-zero.

EDIT: I guess it probably has something to do with p.52 of Protter, Stochastic Integration and Differential Equations, namely where a semimartingale $X_t$ is defined as any adapted process such that it is cadlag and for which the integral of the simple predictable process $H_t$ $$I_X(H)= H_0X_0 + \Sigma_{i=1}^n H_i [X_{T_{i+1}}-X_{T_i}]$$ is continuous.

However, what I don't understand:

1. Why cadlag?

I assume because it guarantees only jump discontinuities, and with that at most countably many discontinuities (like for a distribution function? or is that only because a distribution function is increasing -- if we don't have that the sample paths are of locally bounded variation anymore, does it still hold that we can only have countably many jump discontinuities?)

2. Why do we care if that integral is continuous?

I know it forms the basis for the Ito integral, but what's wrong with having it be a discontinuous process? (If for example we can generalize from continuous semimartingales to arbitrary semimartingales which must be cadlag, why can't we consider cadlag integrals?)

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Chill2Macht
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    What's continuous here is the map $H\mapsto I_X(H)$ (in a certain sense, detailed by Protter), which is an entirely natural thing to ask of an integral. – John Dawkins May 23 '16 at 22:32
  • Oh so it's not a requirement on the indefinite integral itself. Why is it "natural" to ask of an integral? I get why it would be convenient, but not much else. – Chill2Macht May 23 '16 at 22:40
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    We need some kind of "stability" in $I_X$, otherwise, if your $H$ change a tiny little bit, your $I_X(H)$ is light years away, such mapping will not be useful. In Protter's, $H$ is measured using the "strongest" topology (uniform convergence), and $J_X(H)$ using the "weakest" topology (convergence in probability), and such $J_X$ will be the "most" general. – Jay.H May 23 '16 at 22:50
  • That makes a lot more sense to me now; thank you so much! – Chill2Macht May 24 '16 at 00:08

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Check out George Lowther's blog: https://almostsure.wordpress.com/2010/01/03/the-stochastic-integral/ Especially Lemma 3, answers your question "why cadlag?"

Jay.H
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  • I have seen his blog before, but did not know about this post -- you are right, this is very helpful, especially Lemma 3! Thank you so much! – Chill2Macht May 23 '16 at 22:12