is integral with respect to finite variation process of finite variation?

Question

is integral with respect to finite variation process of finite variation?

$\int_{[0,t]}X_sdA_s$, where X is $\mathcal{B}\times F$-measurable.

If no in general, under what conditions?

Question 2: if I have integral w.r. to local martingale. What is the most general process (integrand) for which I can define this integral?

I guess the most general integrand should be only predictable.

Answer 1

For your first question:

$\int_0^t X_s d A_s$ is always of finite variation. For this write $A = A^1 - A^2$ where $A^1, A^2$ are monotone increasing. This possible because $A$ is of finite variation. Now split the integrand in its positive and negative part and rearrange:

$$ \int_0^t X_s d A_s = \int_0^t X_s^+ dA_s^1 + \int_0^t X_s^- dA_s^2 - ( \int_0^t X_s^+ dA_s^2 + \int_0^t X_s^- dA_s^1) = I^1_t - I^2_t $$

which is the difference of two monotone functions and therefore of finite variation.

Answer 2

@NiU and for your first question: I don't see why your statement shouldn't hold. But there is a counterexample of Protter (Stochastic Integration and Differential Equations, 2003) on page 241 (Chapter IV, Ex. 43) which shows that the space of FV-processes is not closed under stochastic integration. Might there be the requirement of the integrand beeing $\mathcal{B}\times F$-measureable violated of the process $H$ in the counterexample of Protter? I don't see this. If your statement is right, it would mean, that for example for any predictable integrand which is integrable by the integrator und FV-process as integrator, the stochastic integral is a FV-process.

For your second question: The most general integrand for your local martingale $X$ is a predictable process which is $X$-integrable. But this class is dependent on $X$, which is probably not what you want. The bigges class of integrands (as far as I know) which is independent of the integrating procces are the predictable local bounded processes. These statements about $X$ hold by the way not only for $X$ beeing local martingale but also for beeing a semimartingale. If you want the local martingale property beeing preserved, you have to require the integrand beeing predictable and locally bounded. Otherwise there is a counterexample of Emery.