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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?

2 Answers 2

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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.

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Since the given SDE is a linear SDE we can solve this equation using a similar approach as for linear ODEs:

  1. Solve the homogeneous SDE $dX_t^\circ = - \lambda \cdot X_t^\circ \, dt$: In this case that's pretty easy, we find $X_t^\circ := X_0^\circ \cdot e^{- \lambda \cdot t}$ as a solution of the homogeneous SDE.
  2. Solve the linear SDE: Choose $X_0^\circ := 1$ and define $Y_t := \frac{1}{X_t^\circ}= e^{\lambda \cdot t}$ (i.e. $\frac{1}{Y_t}$ solves the homogeneous equation for inital value $Y_0=1$). Now set $Z_t := X_t \cdot Y_t$. By applying Itô's formula to $f(x,y) := x \cdot y$ we obtain $dZ_t = Y_t \, dW_t = e^{\lambda \cdot t} \, dW_t \\ \Rightarrow Z_t = Z_0 +\int_0^t e^{\lambda \cdot s} \, dW_s$ Thus $X_t = \frac{Z_t}{Y_t} = e^{-\lambda \cdot t} \cdot \bigg(\underbrace{Z_0}_{X_0 \cdot Y_0=X_0}+ \int_0^t e^{\lambda \cdot s} \, dW_s \bigg)$

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.