= Solution
Define a <probability measure> on the same full measurable space by
$$
\boxed{\frac{d\mathbb P^T}{d\mathbb P}=Z_T=\exp(\mu B_T-\mu^2T/2).}
$$
The <Radon-Nikodym derivative> is positive and has <expectation> one, so this is a <probability measure> and even equivalent to $\mathbb P$, in particular $\mathbb P^T\ll\mathbb P$. For $t\leq T$ its density on $\mathcal F_t$ is $\mathbb E[Z_T\mid\mathcal F_t]=Z_t$.
The finite-horizon <Girsanov theorem> says that if $W$ is a <Brownian motion> and $\mathcal E(\int h\,dW)$ is a true <martingale> through $T$ defining the density, then $W_t-\int_0^th_sds$ is a <Brownian motion> under that density for $t\leq T$. Here $h_s=\mu$, and the <stochastic exponential> is exactly $Z$. The previous part verifies the necessary <martingale> property; alternatively the <Novikov condition> is $e^{\mu^2T/2}<\infty$. Hence $B_t-\mu t$ is a standard <Brownian motion> on $[0,T]$ relative to the same <filtration> under $\mathbb P^T$.
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