Suppose we're given a positive continuous semimartingale $X$ with standard decomposition $X=X_0+A+M$, where $A$ is continuous with paths of bounded variation on finite intervals and where $M$ is a local martingale, with $A_0=M_0=0$ a.s. Now I'm asked for a multiplicative decomposition $X_t=X_0\tilde A_t\tilde M_t$ where $\tilde A_0=\tilde M_0=1$, and where where $\tilde A$ is continuous with paths of bounded variation on finite intervals and where $\tilde M$ is a local martingale.
This can be solved using Itô's formula applied to $\log X$, yielding $$\tilde A=\exp\left(\int_0^{\cdot}\frac1X\ \mathrm dA\right)$$ and $$\tilde M=\mathcal E\left(\int_0^\cdot\frac1X\ \mathrm dM\right)=\exp\left(\int_0^{\cdot}\frac1X\ \mathrm dM-\frac12\int_0^{\cdot}\frac1{X^2}\ \mathrm d\langle M\rangle\right),$$ i.e. $\tilde M$ equals the Doléans exponential of $\int_0^\cdot\frac1X\ \mathrm dX$.
My question: is this decomposition unique? If so, why? I'm aware of the fact that the standard semimartingale decomposition $X=X_0+A+M$ is unique.
Best Answer
Suppose $X_t=X_0A_tM_t=X_0A'_tM'_t$, where $A,M,A',M'$ satisfy the given conditions. Then \begin{align} \log X_t &= \log X_0 + \log A_t + \log M_t\\ &= \log X_0 + \log A_t + \int_0^t \frac1{M_s}\,dM_s - \frac12\int_0^t \frac1{M_s^2}\,d\langle M \rangle_s. \end{align} Similarly, $$ \log X_t = \log X_0 + \log A'_t + \int_0^t \frac1{M'_s}\,dM'_s - \frac12\int_0^t \frac1{(M'_s)^2}\,d\langle M' \rangle_s. $$ Thus, the local martingale parts of these decompositions agree: $$ \int_0^t \frac1{M_s}\,dM_s = \int_0^t \frac1{M'_s}\,dM'_s. $$ The quadratic variation of the above local martingale is therefore $$ \int_0^t \frac1{M_s^2}\,d\langle M \rangle_s = \int_0^t \frac1{(M'_s)^2}\,d\langle M' \rangle_s. $$ Together these show that $\log M_t = \log M'_t$, which implies $M = M'$, and hence, $A = A'$.