On Girsanov's Theorem

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.

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

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.