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Question: Is $X_t = e^{\int_0^t B_sdB_s - \frac{1}{2}\int_0^t B_s^2 dB_s}$ a martingale with respect to the filtration generated by $B_t$?

In order to determine whether the above expression is a martingale, I thought that it'd be a good first step to calculate each of the integrals in the exponent.

Thus, I first found that $\displaystyle \int_0^tB_sdB_s=\frac{B_t^2}{2}-\frac{t}{2}$ by using Ito's lemma on $B_s^2$ and manipulating the resulting equation.

However, I am not sure how to solve for the second integral, $ \displaystyle \int_0^tB_s^2dB_s$. I can't think of an expression on which to utilize Ito's lemma to be able to isolate for the given integral.

Am I even going about this question the right way? Do I need to solve for the integrals? Any guidance would be appreciated.

Jess
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    Computing the integrals in the exponent may not be necessary, at least not right away. Let $X_t = \int_0^t (B_s - \frac{1}{2} B_s^2),dB_s$ be the process in the exponent, and use Ito's lemma to express $f(X_t)$ where $f(x) = e^x$. If you see a bounded variation term, you definitely don't have a martingale. Otherwise, you have a local martingale and can start investigating whether it is a martingale. – Nate Eldredge Mar 09 '15 at 01:47

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