I would like to obtain an explicit solution to $X$ when it satisfies

$$dX_t = \mu X_t dt + \sigma dW_t, X_S = x$$

Here, $S > 0$, and we want an explicit solution for $X_T$, $T > S$.

I am not sure how to approach the problem. Seems more difficult than regular brownian motion!

  • 3
    $\begingroup$ I am voting this question as off-topic for being too basic. Hint 1: Compute the differential of $Y_t = X_t e^{-\mu t}$. Hint 2: This is a special case of an Ornstein-Uhlenbeck process - search a bit and you'll find multiple related questions. $\endgroup$ Commented Feb 26, 2017 at 14:47

1 Answer 1


To solve the SDE you should use the so called variation of constant method. Define a process $Y_t=e^{-\mu t}X_t$, so that using Itô we obtain: $$dY_t=-\mu Y_t dt+ e^{-\mu t}X_t=e^{-\mu t} \sigma dW_t $$ Therefore by integrating we have: $$Y_T=Y_S+\int_S^T e^{-\mu t} \sigma dW_t=e^{-\mu S} X_S +\sigma \int_S^T e^{-\mu t} dW_t$$ $$\Rightarrow \quad X_T=e^{\mu (T-S)} X_S +\sigma \int_S^T e^{\mu (T-t)} dW_t$$ This is a simple Ornstein Uhlenbeck process with mean reversion towards 0 if the coefficient $\mu$ is negative. I think you can easily find it in your Stochastic Calculus reference book.

  • $\begingroup$ The integration limits are wrong, no? $\endgroup$
    – Jaood
    Commented Feb 28, 2017 at 14:32
  • $\begingroup$ Yes you were right. I fixed it $\endgroup$
    – NSZ
    Commented Mar 1, 2017 at 11:07
  • $\begingroup$ @nsz, can you maybe explain how you got the dynamics of $Y_t$ $\endgroup$ Commented May 5, 2021 at 14:06

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