I need help to explain reasoning behind why we choose certain substitutions to solve SDE. After choosing the correct substitution the solution of the SDE are often quite easy. However I have trouble understanding why we choose certain substitutions.

For example

1. Solve the following differential equation:

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

Question 1: Here we set $Y_t = e^{\alpha t}X_t$, find $dY_t$ and then solving the rest of the SDE is easy. Why do we set $Y_t = e^{\alpha t}X_t$? and how do we know how to choose it?

2. and the following: $dX_t = [1-\ln(X_t)]X_tdt + \sigma X_t dB_t$

Here we first set $Y_t = \ln(X_t)$ find

$dY_t = (1-\frac{\sigma^2}{2}-Y_t)dt + \sigma dB_t$

then we need to set $Z_t = e^t Y_t$ and solve:

$dZ_t = e^t(1 - \frac{\sigma^2}{2})dt + e^t\sigma dB_t$

Integrate $dZ_t$ to find $Z_t$ and lastly put it back to $X_t$ giving

$X_t = e^{Y_t} = e^{e^{-t}Z_t}$

Question 2: Again how do we know which substitutions to choose?

Question 3: Is there any general rule for how to set the substitutions?



1 Answer 1


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

Forget a moment about the SDE and consider the associated ordinary differential equation

$$dx_t = (\beta- \alpha x_t) \, dt \tag{1}$$

instead. If I would ask you to solve this ODE, you would (hopefully)

  • first solve the homogeneous equation $$dx_t = -\alpha x_t \, dt$$ and find that the solution of this equation equals $x_t = c e^{-\alpha t}$,

  • use variation of constants to solve the inhomogeneous equation $(1)$, i.e. consider $$x(t) = c(t) e^{-\alpha t}$$ and determine $c(t)$ by plugging $x(t)$ into $(1)$.

Now, since we are interested in the solution to the SDE, we modify the variation of constants approach: we let $c(t)$ depend on $\omega$, i.e. we consider $(c(t))_{t \geq 0}$ as a stochastic process. It remains to determine

$$c(t) = e^{\alpha t} X_t,$$

and to this end, we can use Itô's formula. Note that $c(t) =Y(t)$ (defined in the opening question).

$dX_t = (1-\ln (X_t)) \, X_t \, dt+ \sigma X_t \, dB_t$

Given an SDE of the form

$$dX_t = b(X_t) \, dt + \sigma(X_t) \, dt,$$

there are sufficient and necessary conditions on $b$ and $\sigma$ for the transformation into a linear SDE

$$dY_t = (\alpha + \beta Y_t) \, dt + (\gamma+\delta Y_t) \, dB_t,$$

see this question and the second part of this answer. Applying the criterion gives the transform $Y_t = \ln(X_t)$. The substitution $Z_t := e^t Y_t$ can be motivated using exactly the same reasoning as in the first part of this answer.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .