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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](https://en.wikipedia.org/wiki/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.2012-11-23
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    @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.2012-11-23

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Itō formula has nothing to do with a probability measure. Let me qualify this statement. While stochastic integration ostensibly needs to be performed with the help of a probability measure, the result of stochastic integration is the same for all equivalent measures. In this sense stochastic integration, and therefore also the Itō formula, is a pathwise concept. It cannot be defined pathwise (in general) but the result is pathwise unique (up to indistinguishability). I hōpe this helps :-).

Edit for Bananach: The result of stochastic integration is the same under all equivalent measures: see Proposition 3.6.20 in Bichteler, Klaus, Stochastic integration with jumps, Encyclopedia of Mathematics and Its Applications. 89. Cambridge: Cambridge University Press. xiii, 501 p. (2002). ZBL1002.60001.

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    Can you give a reference for the fact that the result is independent of measure changes? I never fully understood in which sense Ito's integral is a pathwise construction and in which sense it isn't2017-08-14
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    Any standard text, like Protter or Jacod and Shiryaev or Bichteler (who actually refers to this aspect in his introduction). It is known that stochastic integral is unique up to indistinguishability, and the latter is the same for all equivalent measures (just like a.s.).2017-08-14
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    IOW, the formula in the question reads $$E(X_t^2)=2E\left(\int_0^tX_sdX_s\right)+t$$ which is true but is only a very weak version of Itō's formula, which states that $$X_t^2=2\int_0^tX_sdX_s+t$$2017-08-14
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    And +1, if only for the ō in "I hōpe this helps"... :-)2017-08-14
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    @Bananach I have dug up a reference and inserted it in the answer above2017-08-15
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It was introduced to me as "the chain rule of Itō calculus" which I think is a quite elegant description, because it explains it's significance and basic applications. If you know how to calculate the derivatives of your functions $f, g$ then you can derive $f \circ g$.

If you know how to derive a stochastic process $W_t$ and a function $h$, then you can use Itō's formula to derive $h(W_t)$.

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As already pointed out, Itō's formula is a type of chain rule. Let $W_t$ be a Brownian motion (with no drift, starting in 0) Naively, you would guess that the following simple chain rule from analysis holds, $$ df(W_t)=f'(W_t)dW_t. $$ However, integrating and taking expectations on both sides would then yield $$ E[f(W_t)]-f(0)=0\quad\forall t\geq 0 $$ (I used that $E[\int_{0}^{t}g(s)dW_s]=0$ for any adapted process $g(s)$. Intuitively, this holds since in the left-endpoint-rule approximations of the integral, each term has zero expectation because $E[W(s)-W(r)]=0$) This cannot be true in general. For example, if $f(x)=x^2$ as in your question, this would imply that for any $t\geq 0$ you have $W_t=0$ almost surely, which clearly does not hold.

The above motivates an extra term in the chain rule. The correct form, $\frac{1}{2}f''(W_t)dt$, of this extra term at least makes sense in the above example: Since the Wiener process spreads out, the expected value $E[f(W_t)]$ should be increasing for small $t$ when the second derivative $f''(W_0)$ is positive, and decreasing when it is negative.

To arrive at the exact form of Itō's formula (still not rigorously though) one can use a formal Taylor expansion, $$ f(W_t)-f(W_0)=f'(W_0)dW_t+\frac{1}{2}f''(W_0)(dW_t)^2+\mathcal{O}(dW_t^3), $$ together with the intuitive insight that "$(dW_t)^2=dt$" (the variance of $W_t-W_0$ is $t$ by definition of the Wiener process) and "$|dW_t|^r=dt^{r/2}$" for $r>2$, i.e. higher moments are negligible as $dt\to 0$ (again, use that $dW_t$ has normal distribution).

Maybe one can paraphrase the last paragraph as follows:

Itō's formula is the chain rule that you get when higher order terms in the Taylor expansion of $f(X_t)$ become important due to excessive variations of $X_t$.

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    Your last statement is wrong, at best you can say $|dW_t|^r=O(|dt|^{r/2})$ as $dW_t$ has a normal distribution $N(0,dt)=\sqrt{dt}·N(0,1)$.2017-08-14
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    You are right, I fixed it.2017-08-14