My question is at the end of the problem statement.

Solve the following stochastic differential equation.

$dX_t = (\beta - \alpha X_t)dt + \sigma dB_t$, $X_0 = x_0$

where $\alpha$, $\beta$, $\sigma$ are constants, and $\alpha > 0$, and $B$ denotes a one-dimensional standard Brownian motion.

The answer is:

$Y_t = e^{\alpha t}X_t$. The Itô fomula appled to Y then gives us that:

$dY_t = \alpha e^{\alpha t}X_t dt + e^{\alpha t}dX_t = e^{\alpha t}\beta dt + e^{\alpha t}\sigma dB_t$

Integrating we obtain

$Y_t = Y_0 + \int_0^t e^{a(t-s)}\beta ds + \int_0^t e^{-\alpha (t-s)}\sigma dB_s$


$X_t = e^{-\alpha t}x_0 + \frac{\beta}{\alpha}(1-e^{\alpha t}) + \int_0^t e^{-a (t-s)\sigma}dB_s)$

So The question is how do we know that we should use:

$Y_t = e^{\alpha t}X_t$

  • 1
    $\begingroup$ Starting at "Integrating we obtain", formulas should be corrected into $$Y_t = Y_0 + \int_0^t e^{as}\beta ds + \int_0^t e^{\alpha s}\sigma dB_s$$ and $$X_t = e^{-\alpha t}x_0 + \frac{\beta}{\alpha}\left(1-e^{-\alpha t}\right) + \sigma\int_0^t e^{-a (t-s)}dB_s.$$ $\endgroup$
    – Did
    Mar 27, 2016 at 11:05
  • $\begingroup$ Simply look at the segment consisting of $dX_t+\alpha X_tdt+\cdots=\cdots$. You have an "integral factor" which is $e^{\alpha t}$ from the usual ODE theory. That's why you would make the guess with that factor $e^{\alpha t}$. $\endgroup$
    – Xuxu
    Oct 6, 2016 at 0:20

1 Answer 1


One of the possible ways is this. Since the coefficient $\sigma$ is just a constant, you can interpret this equation as a Stratonovich one. Moreover, you can think of it as of an (non-homogeneous) ODE $$\dot{X}_t = \beta - \alpha X_t + \sigma \xi_t,\tag{1}$$ where $\xi_t = \frac{dB_t}{dt}$ is the Gaussian white noise. You can solve the latter using any standard technique from ODE analysis, say, the method of undetermined coefficients.

In other words, if you solve the corresponding homogeneous ODE $$\dot{X}_t = - \alpha X_t, \tag{2}$$ you will get $$X_t = Y e^{- \alpha t}, \tag{3}$$ where $Y$ is a constant. Following the procedure of the method of undetermined coefficients, you can find the solution of the original problem (1) in the form $$X_t = Y_t e^{- \alpha t}, \tag{4}$$ where $Y_t$ is some function of time, and that is how you get to your $X_t = Y_t e^{- \alpha t}$.

P.S. It is possible to clear up the approach mentioned and make it mathematically rigorous using a notion of generalized stochastic processes, however I recommend you look at all of this just as at good heuristics.


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