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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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1 Answer

up vote 1 down vote accepted

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 $L_t$.)

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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
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