Brownian motion relation

Question

How to prove that $P(B_{t_1}^{(\mu)} \in A_1, \dots, B_{t_n}^{(\mu)} \in A_n) = E[1_{B_{t_1} \in A_1, \dots, B_{t_n} \in A_n}e^{\mu B_T}]e^{\frac{-\mu^2 T}{2}}$, considering that $B_t^{(\mu)} = B_t + \mu t, t \geq 0$ is brownian motion with drift?

Answer 1

You have

$$P(B_{t_1}^{(\mu)} \in A_1, \dots, B_{t_n}^{(\mu)} \in A_n)=\int_{A_1 \times \cdots A_n}p^{\mu}(x_1,t_1)p(x_2-x_1,t_2-t_1)\ldots p^{\mu}(x_{t_{n}}-x_{t_{n-1}},t_n-t_{n-1})dx_1\ldots dx_n$$

Here $p^{\mu}$ is the transition density of $B^{\mu}$. Using the definition of a Radon-Nikodym derivative

$$P(B_{t_1}^{(\mu)} \in A_1, \dots, B_{t_n}^{(\mu)} \in A_n)=\int_{A_1 \times \cdots A_n}Z^{\mu}(t_1)p(x_1,t_1)Z^{\mu}(t_1,t_2)p(x_2-x_1,t_2-t_1)\ldots Z^{\mu}(t_{n-1},t_{n})p(x_{t_n}-x_{t_{n-1}},t_n-t_{n-1})dx_1\ldots dx_n$$

Here $p$ is the transition density of $B$. Using standard arguments you should find

$Z^{\mu}(t_{n-1},t_{n})=exp(\mu(B_{t{_n}}-B_{t{_{n-1}}})-\mu^2(t{_n}-t{_{n-1}}))$.

You can see the arguments cancel so that

$Z^{\mu}(t_1)Z^{\mu}(t_1,t_2)\ldots Z^{\mu}(t_{n-1},t_{n})=Z^{\mu}(t_n)=exp(\mu B_{t{_n}}-\mu^2t{_n})$.

From which the rest follows easily. Assuming $t_0=0$, $T=t_n$ otherwise $T=t_n-t_{0}$.