Suppose we have continuous local martingal $L$ given. We define $Z=\mathcal{E}(L)$, the stochastic exponential of $L$. I am interested in finding some condition such that $Z$ defines a density, i.e. I can define an equivalent measure $Q$ by setting $dQ:=Z_\infty dP$.
First of all: $Z$ is again a continuous local martingale, positive hence a supermartingale and $>0$ on $[0,\infty)$. Hence $Z_\infty$ exists. Now I need two assumption, such that everything works out nicely
- $E[Z_\infty]=1$, which is equivalent to the property that $Z$ is a $P$-martingale on $[0,\infty]$, hence uniformly integrable.
- I have to guarantee that $Z_\infty>0$ $P$-a.s.
For $2.$ we have $Z_\infty =e^{L_\infty -\frac{1}{2}\langle L\rangle_\infty}$. Therefore $L_\infty -\frac{1}{2}\langle L\rangle_\infty \not= -\infty$. Since the bracket process is increasing, we must have $\langle L\rangle_\infty <\infty$. The question is:
Can it happen that $\langle L\rangle_\infty <\infty$ but $L_\infty = -\infty$. If so, are there conditions which excludes $L_\infty=-\infty$ and $\langle L\rangle_\infty =\infty$?