Assume we have a standard Brownian motion $W_t$ as the solution to the following SDE

$dX_t=\mu dt+\varepsilon dW_t$

  1. Which kind of SDE it is? Ito process? For which kind of processes we can say that the drift coefficient and $\varepsilon$ does not depend on time or $X$? I assume that in the standard Brownian motion the increments are independent and do not depend on time. So, why in the literature mostly the coefficients depend on time and $X$?

  2. How can i relate this SDE to the Langevin equation governed on the movement of a massless particle immersed in a flow(Brownian motion)

$0=-\gamma \dot x+f(t)$

  • 1
    $\begingroup$ Such $X_t$ is sometimes also called "Arithmetic Brownian Motion". $\endgroup$
    – fes
    Aug 15, 2020 at 11:34

1 Answer 1


Your question is rather unclear. First, the Brownian motion doesn't solve $$\,\mathrm d X_t=\mu\,\mathrm d t+\varepsilon \,\mathrm d W_t,\quad \mu\in\mathbb R, \varepsilon >0.$$ A solution of such equation is called Brownian motion with drift. Moreover, you can find a closed form of the solution : typically $$X_t=X_0+\mu t+\varepsilon W_t,\quad t\geq 0.$$

Notice that a stochastic process is an "Itô process" if and only if it solve an SDE. But an SDE is not an Itô process.

  • $\begingroup$ Thanks. Actually i am more interested to see how one can model the movement of a Brownian particle using an SDE. I think that the Langevin equation is a physical model for that. It seems that the stochastic models ignore the velocity(or mass) in the Langevin equation. It is not clear for me how a SDE is derived starting from the Langevin equation(if there is any relation...) $\endgroup$
    – Denis
    Aug 15, 2020 at 12:15

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