I'd like an explicit formula as a function of $W_t$ (standard Brownian motion) and $\lambda >0$ for the solution of the following SDE:
$\mathrm dX_t = \mathrm dW_t - \lambda X_t \,\mathrm dt$
Someone could help me please?
I'd like an explicit formula as a function of $W_t$ (standard Brownian motion) and $\lambda >0$ for the solution of the following SDE:
$\mathrm dX_t = \mathrm dW_t - \lambda X_t \,\mathrm dt$
Someone could help me please?
The SDE $ \mathrm{d} X_t = - \lambda X_t \,\mathrm{d} t + \mathrm{d} W_t $ has as solution $ X_t = \mathrm{e}^{- \lambda t} X_0 + \int_0^t \mathrm{e}^{- \lambda \left( t - s \right)} \,\mathrm{d} W_s $
If you make the transformation $Y_t = \mathrm{e}^{\lambda t} X_t$, then by Ito's formula applied to $Y_t = f \left( t, X_t \right)$ and after simplifications you get $ Y_t = Y_0 + \int_0^t \mathrm{e}^{\lambda s} \,\mathrm{d} W_s $
By the way, and as additional information, the solution is a special case of the Ornstein-Uhlenbeck process.
Since the given SDE is a linear SDE we can solve this equation using a similar approach as for linear ODEs:
Using this approach one can solve any linear SDE of the form
$dX_t = (\alpha(t)+\beta(t) \cdot X_t) \, dB_t + (\gamma(t) + \delta(t) \cdot X_t) \, dt$
where $\alpha,\beta,\gamma,\delta: [0,\infty) \to \mathbb{R}$ are determinstic coefficients.