# Brownian motion and Girsanov theorem

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 $$T_{a,b}=\inf\{t\geq0:B_t $$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 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

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?

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.

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

• Yes, it was probably a typo. Thank you, I think I understand the connection now. Commented May 4, 2023 at 13:05