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When we apply Ito's Formula to a continuous semimartingale, which is the bounded variation part and which is the continuous local martingale part? Is there a general rule or does it depend on the case?

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If $(X_t)_{t\geq 0}$ is a continuous semimartingale, then it has a representation of the form $$ X_t=X_0+M_t+A_t,\quad t\geq 0, $$ where $X_0$ is $\mathcal{F}_0$-measurable, $(M_t)_{t\geq 0}$ is a continuous local martingale and $(A_t)_{t\geq 0}$ is continuous and of bounded variation. When applying Itô's formula, the continuous local martingale part is everything that involves integration with respect to $(M_t)_{t\geq 0}$ and the bounded variation part is the rest.

In other words, if $f$ is a $C^2(\mathbb{R})$ function, then $(f(X_t))_{t\geq 0}$ and $(f(X_0)+M^f_t+A^f_t)_{t\geq 0}$ are indistinguishable, where $$ M^f_t=\int_0^t\frac{\partial}{\partial x}f(X_s)\,\mathrm dM_s,\quad t\geq 0 $$ is the continuous locale martingale part and $$ A^f_t=\int_0^t\frac{\partial}{\partial x}f(X_s)\,\mathrm d A_s+\frac{1}{2}\int_0^t\frac{\partial^2}{\partial x^2}f(X_s)\,\mathrm d [X]_s,\quad t\geq 0 $$ is the bounded variation part. I'm sure you can extend this to the case where $f\in C^2(\mathbb{R}^n)$.

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And what about the $f(X_0)$ term? Is it included in the bounded variation or the continuous local part? – Nick Papadopoulos Nov 23 '12 at 10:50
It doesn't matter, it it just an $\mathcal{F}_0$-measurable variable. Adding it to a continuous locale martingale still yields a continuous locale martingale, and the same applies for the bounded variation part. But note that, by including it into the locale martingale changes the initial value, i.e. $M_0^f\neq 0$. – Stefan Hansen Nov 23 '12 at 10:53
So, if we "normalize" the resulting semimartingale and demand that the continuous local martingale starts at zero, then we have only one choice for $f(X_0)$ and that is to include it in the bounded variation part $A^f_t$. – Nick Papadopoulos Nov 23 '12 at 10:57
Yes. Clearly it is of bounded variation and continuous because it does not depend on $t$. – Stefan Hansen Nov 23 '12 at 11:06

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