# Exponential martingale and change of measure

$\newcommand{\qq}{\mathbb{Q}}\newcommand{\ee}{\mathbb{E}}$

Denote $Z_t= \exp( \theta B_t - \frac{1}{2}\theta^2t )$

Given the probability measure $\qq(A) := \ee[ Z_t \mathbb{1}_A ]$

I must calcuate $\ee^\qq[B_t]$ where $B_t$ is a Brownian motion.

Would it be $\ee[B_tZ_t]?$

$d( B_tZ_t ) = Z_tdB_t + B_tdZ_t + d[Z,B]_t$

$d[Z,B]_t = dZ_t dB_t=\theta Z_t dt$

So

$Z_t B_t = \int_0^t Z_sdB_s+\theta \int_0^t B_s Z_s dB_s + \theta\int_0^tZ_sds$

$\ee[Z_t B_t] = \theta\int_0^t \ee[Z_s]ds$

$Z$ is a martingale so $\ee[Z_s] = 1$

$\ee^\qq[B_t] = \theta t$

The question is: is this a Brownian motion under $\qq$? I thought it is not because simply because the expectation should be zero. Is this correct, is there any fault in my reason before this point?

I guess $\mu$ is a typo and should be $\theta$ and that $Z_t$ is the exponential in your very first expression and that $B_t$ is a standard Brownian (volatility parameter equal to one)? If so, what you will get with the measure change is called a Brownian motion with drift. The drift would be $\theta t$ without the $1/2$. More information can be found by browsing for Girsanov's theorem.
You have computed $$E^Q(B_t)$$ correctly, and also conclude correctly that $$B_t$$ is hence not a Brownian motion under $$Q$$.
$$B_t-\theta t$$ has mean zero. In fact, this process is a Brownian motion under $$Q$$. You can see this by Girsanov's theorem (which tells you that measure changes of the type you suggested simply add a drift of $$\int_0^t\theta_sds$$ to an otherwise preserved Brownian motion under the new measure), or by Levy's characterization of Brownian motion (a continuous martingale starting at 0 with quadratic variation equal to time elapsed is a Brownian motion).