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As nobody seems to be able to give any kind of answer to that problem over there at math.stackexchange I post this question here:

How can I show with a heuristic argument based on a Taylor expansion that for Stratonovich stochastic calculus the chain rule takes the form of the classical (Newtonian) one?

The intuition goes like this: Concerning Ito calculus the fact that dX^2 = dt results via a Taylor expansion in Ito's lemma - this fact should stay the same with Stratonovich but it should somehow cancel out in there - I just don't know how...

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Hi,

Well you can have a look at the book of Kloeden and Platen "Numerical Solution of Stochastic Differential Equations" where the derivation of Taylor expansion for diffusion is derived based on iterated Wiener Itô (or Stratanovitch) Integrals.

Best Regards

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  • $\begingroup$ @The Bridge: Thank you for the reference. I have the book in front of me and had a cursory look at it... could you please provide me with the exact place where you found the reference for the above question - Thank you again! $\endgroup$
    – vonjd
    Aug 11, 2010 at 11:29
  • $\begingroup$ Well it is hard to answer this the derivation of Stratanovitch-Taylor expansion in this book is quite long. But in the end what you get is a classicla Taylor formula so it matches what was required in your question ins't it ? $\endgroup$
    – The Bridge
    Aug 14, 2010 at 16:12
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I have published a pedagogical paper which, among other things, contains the Taylor expansion for the Stratonovich case (p. 14):

$$f(x_{t+1})\approx f(x_t)+f' \left(x_t+\frac{x_{t+1}-x_t}{2}\right)(x_{t+1}-x_t)$$ $$=f(x_t)+f' \left(\frac{x_{t+1}+x_t}{2}\right)(x_{t+1}-x_t)$$ For the integral notation, we have to make clear that we are now talking about a different integral. To that end we chose a slightly different notation with a circle in front of the integrator. This circle makes it clear that we use the mid-point for the Stratonovich integral: $$f(X_T)-f(X_0)=\int_0^Tf'(X_t) \circ dX_t$$ As one can see the second term disappears in this case. The reason for this is that when one takes the right-hand point one gets a result like the Ito integral but with a negative correction term. The Stratonovich integral is just the arithmetic mean of the two and therefore loses the correction term. And when one compares this in differential notation to the classic chain rule one can see, that we have really achieved what we wanted: they both have the same form: $$df(X_t)=f' \left(X_{t+\frac{1}{2}} \right)dX_t$$ (Strictly speaking this formula would not make sense since it would not be clear what was meant by $t+\frac{1}{2}$ as a subscript, so it is given here for illustrative purposes only.)

More details can be found in the paper:
von Jouanne-Diedrich, Holger, Ito, Stratonovich and Friends (May 18, 2017). Available at SSRN: https://ssrn.com/abstract=2956257 or http://dx.doi.org/10.2139/ssrn.2956257

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