I am trying to prove that the solution of the SDE:
$$
Z_t = Y_t + \int_0^t Z_s dX_s
$$
is:
$$
Z_t = \mathcal{E}(X)_t \left(y_0 + \int_0^t \mathcal{E}(X_s)^{-1}dY_s – \int_0^t \mathcal{E}(X)_s^{-1}d\langle X,Y\rangle_s\right)
$$
where $\mathcal{E}(X)_t := \exp(X_t – \frac{1}{2}\langle X\rangle_t)$ is the stochastic exponential, and $y_0$ is the initial value of Y, and the initial value of $X$ is 0.
So far I have shown, by Ito's formula applied to $Y_t \mathcal{E}(X)_t^{-1}$, that:
$$
Y_t = \mathcal{E}(X)_t\left( y_0 + \int_0^t \mathcal{E}(X)_s^{-1}dY_s – \int_0^t \mathcal{E}(X)_s^{-1}d\langle X,Y\rangle_s – \int_0^t Y_s\mathcal{E}(X)_s^{-1}dX_s + \int_0^t Y_s \mathcal{E}(X)_s^{-1} d\langle X\rangle_s\right)
$$
which feels like it's going in the right direction as it's beginning to resemble the solution, but I don't know how to finish the proof.
Furthermore, I want to prove uniqueness of the solution after this, but I am not sure how I would do this?
Best Answer
Writing \begin{align} E_t&:=\mathcal{E}(X)_t\,,\\ F_t&:=y_0 + \int_0^t \mathcal{E}(X_s)^{-1}dY_s - \int_0^t \mathcal{E}(X)_s^{-1}d\langle X,Y\rangle_s \end{align} Applying Ito to $Z_t=E_tF_t$ gives \begin{align} dZ_t&=E_t\,dF_t+F_t\,dE_t+d\langle E,F\rangle_t\\[3mm] &=dY_t-d\langle X,Y\rangle_t+F_t\,E_t\,dX_t+d\langle X,Y\rangle_t\\[3mm] &=dY_t+Z_\,dX_t\, \end{align} as expected. Here I used \begin{align} dF_t&=E_t^{-1}\,dY_t-E_t^{-1}\,d\langle X,Y\rangle_t\,,\\[3mm] dE_t&=E_t\,dX_t\,,\\[3mm] d\langle E,F\rangle_t&=E_t E_t^{-1}\,d\langle X,Y\rangle_t=d\langle X,Y\rangle_t\,,\\[3mm] d\langle X,\langle X,Y\rangle\rangle_t&=0\,. \end{align} Regarding uniqueness: Let $Z$ and $Z'$ be solutions with $Z_0=Z'_0$. Then $\widetilde{Z}:=Z-Z'$ is a solution of $$ d\widetilde{Z}_t=\widetilde{Z}_t\,dX_t $$ with $\widetilde{Z}_0=0\,.$ Since this equation has a unique strong solution of the form $$ \widetilde{Z}_t=\widetilde{Z}_0\,e^{X_t-\frac{1}{2}\langle X\rangle_t} $$ it follows that $\widetilde{Z}_t\equiv 0$.