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Assume we have processes $A,B,C$ s.t. $A=\int_0^tB_udC_u$ where $C$ is of bounded variation. Now, there is a notation to write this equation as $dA_t=B_tdC_t$. I saw somewhere that one can now go on to define B as $B_t=\frac{dA_t}{dB_t}$. How is this justified and what does $\frac{dA_t}{dC_t}$ mean?

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    I suppose you mean $B_t = \frac{dA_t}{dC_t}$...? – 2017-01-25
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    @saz Yes. Thanks. – 2017-01-25
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    Well, if you divide $$dA_t = B_t \, dC_t$$ (formally) by $dC_t$ you get $$B_t = \frac{dA_t}{dC_t}.$$ It means exactly what is written in the first line of your question: $$B_t = \frac{dA_t}{dC_t}$$ iff $$A_t-A_0 = \int_0^t B_u \, dC_u.$$ – 2017-01-25
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    @saz What i get is that $dC_t(\omega)$ can be identified with a signed measure and i learned that $dA_t=B_tdC_t$ is just a notation for the integral of B w.r.t C. So how is it justified to simply divide here? – 2017-01-25
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    If a measure $\nu$ is absolutely continuous with respect to a measure $\mu$, then $\frac{d\nu}{d\mu}$ is a standard notation for the density of $\nu$ with respect to $\mu$ (which exists by Radon-Nikodym). For this reason it is kind of natural to write $B_t = \frac{dC_t}{dA_t}$ because it is the analogon for signed measures. (And obviously you can't divide by differentials; it's just a formal thing which gives you some intuition why this notion is used.) – 2017-01-25

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