# A question related to Novikov's condition

The well-known 'Novikov condition' says:

Let $L = (L_t)_{t \geq 0}$ be a continuous local martingale null at 0 and $Z = \exp(L - \frac{1}{2} \langle L \rangle)$ its stochastic exponential.

If

$E[\exp(\frac{1}{2} \langle L \rangle_\infty)] \ < + \infty$,

then $Z$ is a (uniformly integrable) martingale on $[0, +\infty]$.

Now to my question:

Is it also true that in this case $Z_\infty > 0, P-a.s.$?

I'm interested in this question, because $Z_\infty > 0$ would ensure that the measure $Q$ defined by $Q[A] := E_P [Z_\infty 1_A]$ is equivalent to $P$.

Thanks for your help! Regards, Si

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Yes because $\mathbb E [L_{\infty},L_{\infty}] < \infty$ will make L an $\mathbb L^2$ bounded martingale, so $L_{\infty}$ is finite (and the limit of $L_t$.)
Aha, do you mean something like: $\mathbb{E}[\exp(\frac{1}{2} \langle L \rangle_\infty] < + \infty$ and $\exp(x) \geq 1 + x$ imply $\mathbb{E}[\langle L \rangle_\infty] < +\infty$ ? (which implies the rest...) Thanks a lot for your help! –  Mad Si Jul 12 '12 at 11:05