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Fix $t>0$, let $B$ be a Brownian motion and let $\sigma$ be a previsible process such that $$\mathbb{E}\left[\text{exp}\left(\frac{1}{2}\int_0^t\sigma_s^2ds\right)\right]<\infty.$$

Then is $$\mathbb{E}\left[\text{exp}\left(\int_0^t\sigma_sdB_s\right)\right]<\infty\qquad?$$

This holds when $\sigma$ and $B$ are independent, but I'm not sure whether it holds in general.

Thoughts

By Novikov's condition we know that, when the above condition holds, the Doleans-Dade exponential is a martingale: $$M_t=\text{exp}\left(\int_0^t\sigma_sdB_s-\frac{1}{2}\int_0^t\sigma_s^2ds\right),$$ but it's not clear to me whether the result follows from this.

[I saw the question Moment generating function of a stochastic integral, but this question appears to be somewhat different.]

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  • $\begingroup$ As far as I remember, it holds that $\mathbb{E}\exp \left( \frac{1}{2} \int_0^t \sigma_s \, dB_s \right)<\infty$, but not necessarily $\mathbb{E}\exp \left( \int_0^t \sigma_s \, dB_s \right)<\infty$. $\endgroup$ – saz Apr 26 '15 at 16:12
  • $\begingroup$ @saz Thanks. Do you have a proof, reference or counterexample? I tried to show $\sigma_s = B_s$ was a counterexample, but didn't get anywhere. $\endgroup$ – user3341907 Apr 26 '15 at 18:06
  • $\begingroup$ The result mentioned in my previous comment is proved e.g. in Revuz & Yor (look for "Novikov condition"). $\endgroup$ – saz Apr 26 '15 at 18:09
  • $\begingroup$ @saz Thanks, I know Novikov's condition. I've edited the question. $\endgroup$ – user3341907 Apr 26 '15 at 19:33

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