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Given $dG_{t}=\alpha S_{t}dt+\upsilon S_{t}dW_{t}$ and $dS(t)={dG_{t}}-\epsilon_{t}dt$. How can I have $S_{t}=\mathbb{E}^{\mathbb{Q}}\left[\int_{t}^{+\infty}e^{-r(s-t)}\epsilon_{t}ds|\mathcal{F}_{t}\right]$ where $W_{t}^{\mathbb{Q}}=W_{t}+\int_{0}^{t}\frac{\alpha-r}{\upsilon}ds$

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Please, specify what is Q. –  Tarasenya Dec 30 '12 at 21:08
    
Tarasenya, $\mathbb Q$ is completely determined by the definition of the brownien motion under $\mathbb Q$,it's a consequense of Girsanov theorem as you can see in my answer. –  Paul Dec 31 '12 at 17:42
    
What is still mysterious is $r$. –  Paul Dec 31 '12 at 17:42
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I didn't understood from where comes $r$ which intervenes at your change of mesure.

We can easelly show by Ito's Lemma that $\forall \lambda \> 0$, the following processs $$ Z_t = \exp \left\{ \int_0 ^t \lambda_s dW_s -\frac {1}{2} \int _0 ^t (\lambda_s )^2 ds \right\}, \ \ t\geq 0$$ is a $\mathcal F_t$- martingal.

So, we can define the probability mesure, equivalent to $\mathbb P$, $ \mathbb Q$ by $$ d \mathbb Q_{\mathcal F_t} = Z _td \mathbb P_{\mathcal F_t}$$

and have a $\mathbb Q $- brownien motion given by $$ W^{\mathbb Q} _ t = W_t -\int_0 ^t \lambda_s ds $$

In your problem $\lambda_s = \frac{r-\alpha}{v} : = \lambda$ , where $r$, $\alpha$ and $v$ must be constants, I suppose. So, $\int_0^t \lambda_s ds = \frac{r-\alpha}{v} t$ and

\begin{align} Z_t &= \exp \left\{ \frac{r-\alpha}{v}W_t -\frac {1}{2} (\frac{r-\alpha}{v} )^2 t \right\}, \ \ t\geq 0 \\ &= \exp \left\{ \lambda W_t -\frac {1}{2} (\lambda)^2 t \right\}, \ \ t\geq 0\end{align}

then, \begin{align} \mathbb E^{\mathbb{Q}} \left \{ \int_0^t e^{-r(s-t)} \epsilon_t ds | \mathcal F _t\right \} &= \mathbb E^{\mathbb{P}} \left \{ Z_t \int_0^t e^{-r(s-t)} \epsilon_t ds | \mathcal F _t\right \} \\ &= \mathbb E^{\mathbb{P}} \left \{ \int_0^t e^{\lambda W_t - \frac{1}{2}\lambda^2 t-r(s-t)} \epsilon_t ds | \mathcal F _t\right \} \\ &= (...)\end{align}

wich should lead to the result using the hypothesys about the dynamics of both $S_t$ and $G_t$. Please verifie also if it's $\mathbb E^{\mathbb{Q}} \left \{ \int_0^t e^{-r(s-t)} \epsilon_s ds | \mathcal F _t\right \}$ or $\mathbb E^{\mathbb{Q}} \left \{ \int_0^t e^{-r(s-t)} \epsilon_t ds | \mathcal F _t\right \}$ as you wrote.

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