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