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Let $B_t$ be a standard Brownian motion under probability $P$. I'm considering computing this: $$E^P\left[ e ^ { B_t + \int_{0}^{t}B_s\mathrm dB_s - \frac{1}{2}\int_{0}^{t}B^2_s\mathrm ds } \right], t\in[0,T]$$ I'm suggested that we should use Girsanov's Theorem, but I am wondering if it works:

Let $Z_t=e ^ {\int_{0}^{t}B_s\mathrm dB_s - \frac{1}{2}\int_{0}^{t}B^2_s\mathrm ds }$, then $E[Z_T]=1$. Let $\mathscr{F}_T=\sigma(B_t,0\leq t \leq T)$, then we can introduce a new measure $\tilde{P}$ by defining as follow: $$ \tilde P(A)=\int_{A}Z_T\mathrm dP, \forall A\in \mathscr{F}_T $$ Obviously $\tilde P$ is also a probability over $\mathscr{F}_T$. Thus $E^{\tilde P}[X]=E^P[XZ_t]$ for $X\in \mathscr{F}_t$.

Then by Girsanov's Theorem, $\tilde{B}_t=B_t-\int_{0}^{t}B_s\mathrm ds$ is a standard Brownian motion under $\tilde P$. Thus we have: $$\begin{align*} E^P\left[ e ^ { B_t + \int_{0}^{t}B_s\mathrm dB_s - \frac{1}{2}\int_{0}^{t}B^2_s\mathrm ds } \right] &= E^P[e^{B_t}Z_t] \\&=E^{\tilde P}[e^{B_t}]\\ &=E^{\tilde P}[e^{\tilde B_t+\int_{0}^{t}B_s\mathrm ds}] \end{align*}$$ Then I don't know what to do next. Is it possible to use Girsanov's Theorem here? Or is there a probabilistic way or pde way to solve this? Thank you!

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    $\begingroup$ Sorry but if $$Z_t=\exp\left(\int_{0}^{t}B_s\mathrm dB_s + \frac{1}{2}\int_{0}^{t}B^2_s\mathrm ds\right)$$ then, for every $t>0$, $$E[Z_t]>1$$ Would you mean $$Z_t=\exp\left(\int_{0}^{t}B_s\mathrm dB_s - \frac{1}{2}\int_{0}^{t}B^2_s\mathrm ds\right)$$ instead? $\endgroup$ – Did Aug 15 '17 at 7:33
  • $\begingroup$ @Did Oh sorry, I made mistake here. Yeah you are right, that should be what you wrote. Thank you! $\endgroup$ – Edward Wang Aug 15 '17 at 14:04
  • $\begingroup$ What is the motivation for calculating this expectation? Is it a problem from a book? $\endgroup$ – saz Aug 15 '17 at 16:10
  • $\begingroup$ @saz I only know how to solve this kind of question using Girsanov's Theorem when instead of Brownian Motion we have deterministic function. So I just make up a question like this. $\endgroup$ – Edward Wang Aug 15 '17 at 16:15
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Yes, this is computable now that you have edited the problem by putting a minus sign in front of the second integral term (rather than a plus sign).

You are on the right track by using Girsanov's theorem. I will use the same change-of-measure and the same notation that you have used in the question.

Since $\tilde B_t = B_t - \int_0^t B_s ds$, it follows from the product rule that that $$\tilde B_t e^{-t} = \frac{d}{dt} \bigg[ e^{-t} \int_0^t B_sds \bigg] \;\;\;\; \Longrightarrow\;\;\;\; \int_0^t \tilde B_se^{-s}ds = e^{-t} \int_0^t B_sds $$ where I integrated both sides to get the implication. Now multiplying both sides by $e^t$ and then differentiating yields \begin{align*} B_t &= \frac{d}{dt} \bigg[ e^t \int_0^t \tilde B_s e^{-s}ds \bigg] \\ &= \tilde B_t +\int_0^t \tilde B_s e^{t-s}ds \\ &= \int_0^t e^{t-s}d \tilde B_s\end{align*}

where I used a stochastic integration-by-parts in the last equality.

Thus we have written $B_t$ in terms of the process $(\tilde B_s)$. But since $(\tilde B_s)$ is a Brownian motion under $\tilde P$, the above computation reveals that $B_t$ is distributed as a mean-zero Gaussian under $\tilde P$. Moreover its variance can be easily computed by using the Itô isometry: $$\sigma^2=\Bbb E^{\tilde P} [ B_t^2 ] = \Bbb E^{\tilde P} \bigg[ \bigg( \int_0^t e^{t-s}d\tilde B_s \bigg)^2 \bigg] = \int_0^t e^{2(t-s)}ds = \frac{1}{2}(e^{2t}-1)$$

Now, if $Z$ is a mean-zero Gaussian, then it is well-known that $E[e^Z] = e^{\frac{1}{2}\sigma^2}$, therefore we find that $$\Bbb E^P\left[ e ^ { B_t + \int_{0}^{t}B_s\mathrm dB_s - \frac{1}{2}\int_{0}^{t}B^2_s\mathrm ds } \right]=\Bbb E^{\tilde P}[e^{B_t}] = e^{\frac{1}{4}(e^{2t}-1)}$$ I may have made an error in some of the computations, maybe someone can post a different method to verify.

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