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I am working through the proof of Itō's formula contained in the book "Continuous Martingales and Brownian Motion" by Revuz and Yor and am stuck at a point in the proof.

Theorem (Itō's formula). Let $X = (X^1, \ldots, X^d)$ be a continuous vector semimartingale and $F \in C^2(\mathbb{R}^d, \mathbb{R})$; then, $F(X)$ is a continuous semimartingale and $$ F(X_t) = F(X_0) + \sum_i \int_0^t \frac{\partial F}{\partial x_i}(X_s)dX_s^i + \frac{1}{2}\sum_{i,j}\int_0^t\frac{\partial^2 F}{\partial x_i\partial x_j}(X_s)d\langle X^i, X^j\rangle_s. $$

The equation immediately shows that $F(X)$ is a continuous semimartingale. Through stopping and approximation it suffices to prove the equation in the case where $F(\cdot)$ is a polynomial. Therefore it is enough to show the following claim, since the equation is true for constant functions.

Claim. If the equality holds for the polynomial $F$, then it holds for all polynomials $$G(x_1, \ldots, x_d) := x_{i_0}F(x_1, \ldots, x_d), \;\; 1 \leq i_0 \leq d$$

Using the integration by parts formula, one obtains $$ G(Y_t) = G(Y_0) + \int_0^t Y_s^{i_0}dF(Y)_s +\int_0^t F(Y_s)dY_s^{i_0} + \langle Y^{i_0}, F(Y)\rangle_t$$

This is now the point where I am not sure how to proceed exactly. Clearly, I want to "replace" the $dF(Y)_s$ in the first integral using Itō's formula for $F$ (which we assumed to hold) to obtain

\begin{align} G(Y_t) = G(Y_0) & + \sum_{i=1}^d \int_0^t Y_s^{i_0}\partial_i F(Y_s) dY_s^i + \int_0^t F(Y_s) dY_s^{i_0} \\ & + \frac{1}{2}\sum_{i,j}^d \int_0^t Y_s^{i_0}\partial_{i,j}^2F(Y_s)d\langle Y^i, Y^j\rangle_s + \langle Y^{i_0}, F(Y) \rangle_t \end{align}

My question now is, how to justify this in a rigourous manner or which intermediate steps I am missing, since it is not obvious to me why one can just make this formal replacement.

Thanks for any input.

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I think what you are missing is the "composition property" (I do not remember the name exactly) of stochastic integrals (this is proposition IV.2.4 in Revuz-Yor, this is the case $L^2$ but it can be later be generalized), if $$ V_t = \int_0^t X_s dY_s $$ where $$ Y_t = \int_0^t W_s dZ_s $$ then $$ V_t = \int_0^t X_s W_s dZ_s$$ This can be reinterpreted in the differential notation as $$ dV_s = X_s W_s dZ_s$$ In your case $Y = F(X) = \text{Ito's formula}$ and $G = V$

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  • $\begingroup$ Thanks, that is exactly what I have been looking for! $\endgroup$ – caligula Jun 27 '13 at 13:28

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