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I have a question concerning a calculus "trick" sometimes used in stochastic calculus (e.g. in the Book on Arbitrage Theory in Cont. Time of Bjoerk). There they do the following in the proof of Prop. 29.6:

$\frac{1}{B_t}dF_t=h_tdW_t$, multiply on both sides with $B_t$ to obtain $dF_t=B_th_tdW_t$. The step seems rather obvious. $B_t$ is a prob. a finite variation process. The question is: Can one apply this also in general?

I.e. $F$ is a solution to

$dM_t = F_td\frac{1}{S_t}$ iff it solves $S_tdM_t = S_tF_t \frac{1}{S_t}$

$M$ depends on F, but is a local martingale. Say $F_td\frac{1}{S_t}$ is a (strict) local martingale. Multiplying $S$ gives $S_t F_td\frac{1}{S_t}$ which might well be a true martingale. So this transformation can have significant impact. Is it allowed?

Thank you for you attention!

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1 Answer 1

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Written out in detail, the statement $\frac{1}{B_t}dF_t = h_tdW_t$ means

$$ \int_0^t \frac{1}{B_s}dF_s = \int_0^t h_s dW_s \qquad\textrm{for all }t\ge0, $$ so that $B^{-1}\cdot F = h\cdot W$, in process notation, where $(h\cdot W)_t = \int_0^t h_sdW_s$, et cetera. Using the composition rule for stochastic integration, namely $K\cdot (H\cdot X) = KH\cdot X$ (see for example Theorem II.19 of Protters book), yields that the above equality implies $F = B \cdot B^{-1} \cdot F = B\cdot (h\cdot W) = Bh\cdot W$, meaning $dF_t = B_th_tdW_t$. Whether $B$ has finite variation is irrelevant, it suffices that $B$ for example is predictable and locally bounded (this is in particular the case if $B$ is continuous). If $B$ is everywhere positive, the other implication holds as well, as we may then simply use the composition rule with $B^{-1}$.

By similar arguments in the general case, the following result holds: Let $X$ and $Y$ be some semimartingales, and let $H$ be some positive, predictable and locally bounded process. It then holds that

$$X = Y \qquad \textrm{if and only if}\qquad H \cdot X = H \cdot Y$$

or, in other words,

$$dX_t = dY_t \qquad \textrm{if and only if}\qquad H_tdX_t = H_tdY_t.$$

Hope this answers your question.

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Well, it does. Thank you. –  user13655 May 16 '13 at 16:46

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