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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?

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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If I am not mistaken, this is the very first tumbleweed question – user285523 Nov 23 '15 at 15:01
@0.5772156649... Indeed, didn't know that :-) – vonjd Nov 23 '15 at 17:40
up vote 6 down vote accepted

In the following, $W_t$ is a Wiener process with increments having mean zero and standard deviation 1.

Basically, the Stratonovich formula does the following in terms of the Itô formula:

$$f(W_t) \circ dW_t = \frac{f(W_t+dW_t)+f(W_t)}{2}dW_t \; ,$$

which can be rewritten as

$$f(W_t) \circ dW_t = \frac{f(W_t+dW_t)-f(W_t)}{2}dW_t + f(W_t)dW_t \; .$$

Using Taylor expansion on the first term and Itô calculus rules, this can be simplified to

$$f(W_t) \circ dW_t = \frac{f'(W_t)}{2}dt + f(W_t)dW_t \; .$$

Now, we can replace $f$ with $f'$ in the formula to give

$$f'(W_t) \circ dW_t = \frac{f''(W_t)}{2}dt + f'(W_t)dW_t \; .$$

But you can easily check that the right hand side is nothing but $df(W_t)$ according to Itô calculus rules, therefore

$$df(W_t) = f'(W_t) \circ dW_t \; .$$

Which is just a special case of the chain rule. I guess from here on out you can generalize the argument.

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