# Compute $\tilde{E}\left(B_t - \int_0^t B_s \,ds\right)$

We have that $$B$$ is a Standard Brownian Motion with $$B_0 = 0$$ under the prob. measure $$P$$ and $$\tilde{B} = B_t - \int_0^t B_s ds$$ for $$t\in [0,T]$$, $$T>0$$. I want to calculate $$\tilde{E}\left(B_t - \int_0^t B_s \,ds\right)$$ where $$\tilde{E}$$ is the expectation under the probability measure given by Girsanov Theorem... My question is: Since $$\tilde{B}$$ is a SBM under $$\tilde{P}$$, is it not zero? Otherwise, how could I do it explicitly? Thanks in advance

• Consider process $X_t = B_t - \int B_s ds$, show that this process is gaussian. Then integrate it using Girsanov measure. Jan 10, 2021 at 23:23
• If $\bar B$ is a standard BM under $\bar P$, then $E_{\bar P} [\bar B_t]=0.$
– UBM
Jan 10, 2021 at 23:26
• @onespace Is this last part of integrating using Girsanov where I'm struggling a bit
– R__
Jan 10, 2021 at 23:27
• @UBM Yeah, I know that it must be zero indeed haha but I think this exercise is written to check that this is true... so they are asking to calculate it explicitly
– R__
Jan 10, 2021 at 23:28
• @openspace One question: is it true that $E(Z(B_t-B_s) = E(Z) E(B_t-B_s)$? I think it's the state I need
– R__
Jan 10, 2021 at 23:30

Given $$\bar B_t = B_t - \int_0^t B_s ds, \tag{1}$$ the corresponding Radon–Nikodym density process of $$Q$$ with respect to $$P$$ is $$\rho_t:=\left. \frac{dQ}{dP} \right|_{\mathcal F_t}=e^{\int_0^t B_s dB_s - \frac{1}{2} \int_0^t B_s^2ds}, \quad 0 \leq t \leq T,$$ which solves the SDE $$\rho_t = \rho_t B_t dB_t, \quad \rho_0 = 1. \tag{2}$$ Now, apply the integration by parts formula (Ito product rule) using (1) and (2) to show that $$d(\bar B_t \rho_t)=A_t \cdot dB_t$$ (for some process $$A$$). This means that we don't have a drift in the SDE of $$B_t \rho_t.$$ Thus, $$\{B_t \rho_t, 0 \leq t \leq T\}$$ is a $$P$$-local martingale. Then check that $$E_P[\int_0^t A^2_s ds] < \infty$$ to conclude that $$\{B_t \rho_t, 0 \leq t \leq T\}$$ is indeed a $$P$$-martingale. Therefore, $$E_Q[\bar B_t] = E_P[\bar B_t \rho_t]= E_P[\bar B_0 \rho_0]=E_P[\bar B_0]=E_P[B_0]=0.$$

• I don't understand why the squared integrbaility of A shows that this is a martingale with respect to $\tilde{P}$ :(
– R__
Jan 11, 2021 at 11:15
• Why do you say $\bar P$? It's martingale with respect to $P.$
– UBM
Jan 11, 2021 at 12:56
• The biggest space for what the Ito integral is defined is the class of adapted process $h$ such that $\int_0^t h^2_u du < \infty.$ The process $I_t:=\{\int_0^t h_udB_u, 0 \leq t \leq T\}$ is a local martingale. If we restrict the class of integrands $h$ to $\{h \text{ adapted such that } E \int_0^t h^2_u du < \infty, \}$ then the process I is a martingale. You will find this in every book where the Ito integral is defined.
– UBM
Jan 11, 2021 at 13:07