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From the point of view of analysis, what is Ito formula?
Is it an integral by substitution, or, a radon-nikodym derivative?

Define the probability space $$ \left(C\left(\Bbb R_+\right),\sigma\left(C\left(\Bbb R_+\right)\right),P\right), $$ where $P$ is the standard Brownian motion measure.
Let $f(x)=x^2$, with Ito formula, I write $$ \int_{C\left(\Bbb R_+\right)} f(X_t)dP(X)=\int_{C\left(\Bbb R_+\right)} \left\{\int_0^t 2X_sdX_s+t \right\} dP\left(X\right). $$ The previous equation is my heuristic to explain Ito formula to an analyst.
This heuristic is itself inspired by the following heuristic which I once heard

Ito formula is a way of expressing how the measure $P$ changes from $X_t$ to $f(X_t)$

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I know it's not easy to enter on most keyboards, but his name is more correctly (Hepburn) romanized as "Itō" with a macron over the 'o' to indicate a long vowel. As a side note it's quite hard for non-native speakers to distinguish 'o' and 'ō' without practise. If you ever need a macron in a LaTeX paper you can use \= for it the same way you would for any other accent, i.e. \=o produces 'ō' - or one could use a unicode-aware TeX engine and just enter 'ō' directly. – kahen Nov 23 '12 at 16:49
    
@kahen When I look at other post, people write 'Ito' with no accent. I guess I can follow this practice, as effectively, the macron is hard to type. Thank you for telling me. I will correct it in my different posts. – Nicolas Essis-Breton Nov 23 '12 at 17:00

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