Assume $X$ is a finite-variation Lévy process and let $Y_t = e^{X_t}$. It can be shown that one can decompose $Y_t$ as $Y_t = M_t + A_t$ where $M$ is a local martingale and $A$ is continuous and finite variation. Am trying to verify whether there exists a predictable process $\lambda$ such that $$ dA_t = d\langle M \rangle_t \lambda_t\tag1 $$ where $\langle M \rangle$ is the predictable quadratic variation process. From what I have found, $A$ and $M$ have the following expressions: $$ A_t = \int^t_0 Y_{s-} \alpha\, \mathrm ds, \qquad \langle M \rangle_t = \int^t_0 Y^2_{s-} \beta\, \mathrm ds $$ for constatnts $\alpha, \beta$. But then, $(1)$ seems to become $$ \alpha = \beta Y_t\lambda_t \Leftrightarrow \lambda_t = \frac{\alpha\beta^{-1}}{Y_t} $$ but $Y_t$ is not predictable, which seems to imply that it does not exist such a predictable process $\lambda_t$ that satisfies $(1)$.
I am pretty sure about the expressions of $A_t$ and $\langle M\rangle$ but I was somewhat surprised by the conclusion which I fear might be incorrect. I would be very grateful if someone could verify whether my analysis is correct.