I am studying on Girsanov's theorem. Moreover my lecturer states, following:
$$Z>0\text{ is a continuous local martingale }\Leftrightarrow \exists! \text{ N continuous local martingale }:Z=e^{(N-\frac{1}{2}[N])}$$
Is this still true, when $N$ is only a martingale?
$$Z>0\text{ is a continuous martingale }\Leftrightarrow \exists! \text{ N continuous martingale }:Z=e^{(N-\frac{1}{2}[N])}$$
If yes, is it possible to rewrite the condition "If the Doléans-Dade exponential is a continuous martingale" into "If $N$ is a continuous martingale" ? (I am especially interested in $"\Leftarrow"$)
Proof for the statement: Using Ito's lemma we find both directions:
$"\Leftarrow"$
$$ \begin{align} Z&=\exp(N-\frac{1}{2}[N])\\ &=\exp(N_0)+\exp(N-\frac{1}{2}[N])\cdot (N-\frac{1}{2}[N])+\frac{1}{2}\exp(N-\frac{1}{2}[N])\cdot[N]\\ &=\exp(N_0)+\exp(N-\frac{1}{2}[N])\cdot N \end{align} $$
$"\Rightarrow"$
$$ \begin{align} \log Z=\log Z_0+Z^{-1}\cdot Z-\frac{1}{2}Z^{-2}\cdot[Z]=(\log Z_0+Z^{-1}\cdot Z)-\frac{1}{2}[\log Z_0+Z^{-1}\cdot Z] \end{align} $$
is a semi-martingale with unique decomposition, therefore $Z=e^{(N-\frac{1}{2}[N])}$ with $N:=\log Z_0+Z^{-1}\cdot Z$
Looking at the proof, my question seems to narrow down to the following: When is a stochastic integral with respect to a continuous martingale a continuous martingale again?