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Let's consider the probability space $ (\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb{P})$ and a $\mathcal F_t$-Brownian motion under $\mathbb{P}$, $(W^{\mathbb P}_t, t\geq 0) $ with $W^{\mathbb P}_0=0$.

We can easily show by Itô's Lemma that $\forall \lambda \> 0$, the following process $$ D^{(\lambda)} _t = \exp \left\{ -\frac{\lambda}{2} (W^{\mathbb P}_t )^2 +\frac{\lambda t}{2}-\frac{\lambda^2}{2} \int _0 ^t (W^{\mathbb P}_s )^2 ds \right\}, \ \ t\geq 0$$ is a $\mathcal F_t$- martingale.

So, we can define the probability measure, equivalent to $\mathbb P$, $ \mathbb Q^{(\lambda)}$ by $$ d \mathbb Q^{(\lambda)}_{\mathcal F_t} = D^{(\lambda)} _td \mathbb P_{\mathcal F_t}$$

also, by the Girsanov theorem we show that there exists a $\mathbb Q^{(\lambda)} $ -Brownian motion, $(W^{\mathbb Q^{(\lambda)}}_t, t\geq 0) $ with $W^{\mathbb Q^{(\lambda)}}_0=0$ and satisfying the SDE :

$$ dW^{\mathbb P}_t= dW^{\mathbb Q^{(\lambda)}}_t -\lambda W^{\mathbb P}_t dt $$

My question: How can we obtain an explicit formula as a function of $W^{\mathbb Q^{(\lambda)}}_t$ for the solution of this SDE?

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1 Answer 1

up vote 1 down vote accepted

Hi Paul unless I missed something in Girsanov's theorem, it should give :

$dW^{\mathbb Q^{(\lambda)}}_t =dW^{\mathbb P}_t - \lambda W^{\mathbb P}_t dt$

or :

$W^{\mathbb Q^{(\lambda)}}_t =W^{\mathbb P}_t - \lambda \int_0^t W^{\mathbb P}_s ds$

And this is an explicit solution given that $W^{\mathbb P}$ is given from the beginning.

Best regards

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