Find SDE satisfied by transformation of solution to a different SDE

Solution 1:

Let $Z_t=\frac{t}{2}+\ln{(X_t)}$ and denote quadratic variation by $\langle\cdot\rangle$. By Ito, $$dZ_t=d\left(\frac{t}{2}+\ln{(X_t)}\right)=\frac{dt}{2}+\frac{dX_t}{X_t}-\frac{\langle X_t\rangle}{X_t^2}\,dW_t$$ Now note that \begin{gather*} X_t=e^{Z_t-\frac{t}{2}} \\ dX_t=Y_t(dt+dW_t) \\ \langle X_t\rangle=Y_t \end{gather*} Thus \begin{align*} dZ_t&=\frac{dt}{2}+Y_te^{\frac{1}{2}t-Z_t}(dt+dW_t)-Y_te^{t-2Z_t}\,dW_t \\ &=\left(\frac{1}{2}+Y_te^{\frac{1}{2}t-Z_t}\right)dt+Y_te^{\frac{1}{2}t-Z_t}\left(1-e^{\frac{1}{2}t-Z_t}\right)\,dW_t \end{align*}

You've already noticed that $Y_t=Y_0e^{W_t-\frac{t}{2}}$, so we can substitute that to get a slightly nicer answer: $$dZ_t=\left(\frac{1}{2}+Y_0e^{W_t-Z_t}\right)dt+e^{W_t-Z_t}\left(1-e^{\frac{1}{2}t-Z_t}\right)\,dW_t$$