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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

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  • $\begingroup$ Consider process $X_t = B_t - \int B_s ds$, show that this process is gaussian. Then integrate it using Girsanov measure. $\endgroup$
    – openspace
    Jan 10, 2021 at 23:23
  • $\begingroup$ If $\bar B$ is a standard BM under $\bar P$, then $E_{\bar P} [\bar B_t]=0.$ $\endgroup$
    – UBM
    Jan 10, 2021 at 23:26
  • $\begingroup$ @onespace Is this last part of integrating using Girsanov where I'm struggling a bit $\endgroup$
    – R__
    Jan 10, 2021 at 23:27
  • $\begingroup$ @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 $\endgroup$
    – R__
    Jan 10, 2021 at 23:28
  • $\begingroup$ @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 $\endgroup$
    – R__
    Jan 10, 2021 at 23:30

1 Answer 1

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

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  • $\begingroup$ I don't understand why the squared integrbaility of A shows that this is a martingale with respect to $\tilde{P} $ :( $\endgroup$
    – R__
    Jan 11, 2021 at 11:15
  • $\begingroup$ Why do you say $\bar P$? It's martingale with respect to $P.$ $\endgroup$
    – UBM
    Jan 11, 2021 at 12:56
  • $\begingroup$ 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. $\endgroup$
    – UBM
    Jan 11, 2021 at 13:07
  • $\begingroup$ This question may help you: math.stackexchange.com/questions/3952224/… $\endgroup$
    – UBM
    Jan 11, 2021 at 20:15
  • $\begingroup$ Now is clear! Thank you very much $\endgroup$
    – R__
    Jan 12, 2021 at 10:57

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