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

Question

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

Answer 1

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, $$E_Q[\bar B_t] = E_P[\bar B_t \rho_t]= E_P[\bar B_0 \rho_0]=E_P[\bar B_0]=E_P[B_0]=0.$$