Showing the Itô logarithm is the inverse of the Doléans-Dade exponential

Question

Consider the stochastic exponential: $F[M] = e^{M(t)-\frac{1}{2}\langle M\rangle(t)}$ for an local martingale $M$.

Define: $$M:= \log(L(0)) + \int_0^* \frac{1}{L} dL $$ where $L$ is a strictly positive local martingale.

How can I apply Ito's lemma to deduce $L= F[M]$ ?

Idea:

$F[M](t)$ yields: $e^{ log(L_0) + \int_0^t \frac{1}{L}(s) dL(s) -0.5\int_0^t \frac{1}{L^2}(s) dL(s) } $

The exponent is just $log(L(t))$. This can be shown by applying Ito's lemma on $L$ with function $f(x) =\log x$

Is this the right way to proof my desired result.

Answer 1

If $L$ is a strictly positive continuous local martingale, then by Ito, $$ \log L_t =\log L_0+\int_0^t L_s^{-1} dL_s-{1\over 2}\int_0^t L^{-2}_sd\langle L\rangle_s. $$ Call the stochastic integral on the right $M_t$, in which case the second integral is $\langle M\rangle_t$. Exponentiating: $$ L_t=L_0\exp\left( M_t-{1\over 2}\langle M\rangle_t\right). $$