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I am trying to use Girsanov's Theorem to relate $dX_t^\mathbb{Q}$ and $dX_t^\mathbb{P}$.

The Girsanov's Theorem I learnt from:

Let $u \in L^2[0,T]$ be a deterministic function where $L^2[0,T] = \{u:[0,T]\to\mathbb{R}; \mathrm{measurable \,and} \int_0^t|u(s)|^2 ds < \infty\}$

Then the process $$X_t = \int_0^tu(s)\,ds +W_t\quad 0\leq t \leq T$$ is a Brownian motion with respect to the probability measure $Q$ given by$$ dQ=e^{-\int_0^Tu(s)\,dW_s-\frac{1}{2}\int_0^Tu(s)^2\,ds}dP$$

However, is it true that $dW_t^\mathbb{Q} = dW_t^\mathbb{P}+u(t)dt$?

$dW_t^\mathbb{Q}, dW_t^\mathbb{P}$ are the differentials of $W_t, W_t$ with respect to measure $\mathbb{Q}, \mathbb{P}$ respectively.

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    $\begingroup$ what means $dX_t^{\mathbb P}$ or $dX_t^{\mathbb Q}=dW_t^{\mathbb P}+...$ ? $\endgroup$
    – Surb
    Oct 16, 2019 at 10:40
  • $\begingroup$ Would $dW_t^\mathbb{Q} = dW_t^\mathbb{P}+u(t)dt$ be true instead? If yes, may you tell me why? $\endgroup$ Oct 16, 2019 at 16:38
  • $\begingroup$ Your notation are unclear for me. $\endgroup$
    – Surb
    Oct 17, 2019 at 7:36

1 Answer 1

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The stochastic differentials $dW^\Bbb Q_t$ and $dW^\Bbb P_t$ are the same, because $\Bbb Q$ and $\Bbb P$ are locally equivalent measures; that is, for $A\in\mathcal F_t$, $\Bbb Q(A) =0$ if and only if $\Bbb P(A)=0$. This is really a statement about stochastic integrals of the form $\int_0^t H_s\,dW_s$ (with respect to $\Bbb Q$ or $\Bbb P$); these integrals are limits in $L^2$ of (identical) Riemann sums. As such, they are also almost sure limits for appropriate subsequences. And since the notion of ``almost surely'' is the same for such sums be it for $\Bbb Q$ or $\Bbb P$, the limits are a.s. the same.

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