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.


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.

  • 1
    $\begingroup$ due to Jordan decomposition: function is of bounded variation if and only if can be written as the difference of two bounded nondecreasing functions. So, I think it should be noticed that $I^1_t, I^2_t$ are bounded functions because $A^1, A^2, X^+, X^-$ are all bounded functions $\endgroup$ – Andrey Pak May 12 '18 at 3:30

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