Brownian motion and Girsanov theorem

Question

I am working on a problem regarding Girsanov theorem and Brownian motion. Let $B$ be a standard Brownian motion. For $a,b>0$ we define stopping times $$ T_{a,b}=\inf\{t\geq0:B_t<bt-a\}$$ $$ S_a=\inf\{t\geq0:B_t=-a\}$$ and a martingale $$ M_t=\exp(B_t-\frac{1}{2}t).$$ Using Girsanov theorem I want to prove the following equality: $$ E(M_t\cdot\mathbb{1}(t<T_{a,1}))=P(t<S_a).$$ I also want to show that $$E(e^{\frac{b^2}{2}T_{a,b}})=e^{ab}.$$

So far I have proven the following two things: $$E(e^{\frac{1}{2}T_{a,1}})\leq e^a$$ $$|M_{T_{a,1}\land t}-M_{T_{a,1}}|\leq M_t\cdot 1(t<T_{a,1})+e^{\frac{1}{2}T_{a,1}}\cdot 1(t<T_{a,1})$$

I am struggling to see the connection between Girsanov theorem and what I want to prove. The equality reminds me of the definition of the new measure $Q$: $Q(A)=E_P(X\cdot1_A)$, where $E(X)=1$, which is true for $X:=M_t$, but as for $A$, $T_{a,1}$ and $S_a$ are not equal. Do I somehow need to prove that $B$ is a Brownian motion with a drift under new $P$? I am not sure how Girsanov theorem proves this equality. Would this equality also help me with the second one I want to prove?

Answer 1

Girsanov theorem tells us that if $B_t$ is a standard Brownian wrt $\mathbb{P}$ then $W_t = B_t - bt$ is a standard Brownian wrt $\mathbb{Q}$ where $\eta_T = \frac{d\mathbb{Q}}{d\mathbb{P}} = \exp(bB_T-b^2T/2)$.

For the first part, observe that almost surely

$$ \{t < T_{a,b}\} = \{\inf_{s \le t} B_s \ge bs - a\} = \{\inf_{s \le t} W_s \ge - a\} \overset{(*)}{=} \{\inf_{s \le t} W_s > - a\} = \{t < S_{a}\} $$

where we take $S_a = \inf\{s: W_s = -a\}$ and (*) is due to continuity of sample paths. Hence, using Girsanov, $\mathbb{Q}(\{t < S_{a}\}) = \mathbb{E}_{\mathbb{P}}(\eta_T ;\{t < T_{a, b}\})$.

For the second part, is there potentially a transcription error? We know $M_t = \exp(b W_t - b^2t/2)$ is a martingale wrt $\mathbb{Q}$. Hence by optional sampling theorem,

$$ 1 = \mathbb{E}_{\mathbb{Q}}(M_{T_{a,b}}) = \mathbb{E}_{\mathbb{Q}}(\exp(b (-a) - b^2T_{a,b}/2)) $$

i.e.

$$ \mathbb{E}_{\mathbb{Q}}(-b^2T_{a,b}/2)) = e^{ab} $$