For SDE's of the general form $$dX_t = b(X_t) \, dt + \sigma(X_t) \, dW_t \tag{1}$$ @saz taught me that there is a formula to transform it into a linear SDE, quoting from René L. Schilling/Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes, p.278.

However I don't have the book. So I'm wondering, is there a name for such formula? so that I could search on Internet or other books for more details or related topics. For example,

  • how is $\alpha$, $\beta$ and $\gamma$ in the formula defined,
  • if the $b(X_t)$ could be more general as $b(t, X_t)$,
  • how to solve the linear SDE after the transformation.

Below are the details of the transformation from @saz's post :

The SDE (1) can be transformed into a linear SDE $$dZ_t = (\alpha+ \beta \cdot Z_t) \, dt + (\gamma+\delta \cdot Z_t) \, dW_t$$ if and only if $$\frac{d}{dx} \left( \frac{\frac{d}{dx}(\kappa'(x) \cdot \sigma(x))}{\kappa'(x)} \right) = 0 \tag{2}$$ where $\kappa(x) := \frac{b(x)}{\sigma(x)}- \frac{1}{2} \sigma'(x)$. The transformation $Z_t = f(X_t)$ is given by $$f(x) := \begin{cases} e^{\delta \cdot d(x)} & \delta \neq 0 \\ \gamma \cdot d(x) & \delta = 0 \end{cases}$$ where $$d(x) := \int_0^x \frac{1}{\sigma(y)} \, dy \qquad \qquad \delta = - \frac{\frac{d}{dx}(\kappa'(x) \cdot \sigma(x))}{\kappa'(x)}$$

  • $\begingroup$ @saz What is $\gamma$? $\endgroup$ – Kerry Nov 7 '15 at 18:37
  • $\begingroup$ @Pii I'm not aware of an explitict formula for $\gamma$. Just make the ansatz $f(x) = \gamma d(x)$ (if $\delta=0$), apply Itô's formula and check whether you can choose suitable $\gamma$ such that the transformation coincides with the original process. $\endgroup$ – saz Nov 8 '15 at 7:16
  1. Since the transformation is given explicitely, you can calculate the coefficients by applying Itô's formula. But I guess, the calculations will be rather tedious and the formulas quite lengthy. If you are given an explitcit SDE of the form $$dX_t = b(X_t) , dt + \sigma(X_t) \, dW_t$$ you obtain the constants by applying the given transformation.
  2. The given formula works only for autonomous coefficients, i.e. coefficients which do not depend on the time. A criterion for non-autonomous coefficients is for example the following:

    The SDE $$dX_t = b(t,X_t) \, dt + \sigma(t,X_t) \, dW_t$$ can be transformed via $Z_t = f(t,X_t)$ into the form $$dZ_t = \bar{b}(t) \, dt + \bar{\sigma}(t) \, dW_t,$$ where $\bar{b}$, $\bar{\sigma}: [0,\infty) \to \mathbb{R}$ are deterministic functions, if and only if $$0 = \frac{\partial}{\partial x} \left( \sigma(t,x) \cdot \left( \frac{\sigma_t(t,x)}{\sigma^2(t,x)} - \frac{\partial}{\partial x} \frac{b(t,x)}{\sigma(t,x)} + \frac{1}{2} \sigma_{xx}(t,x) \right) \right)$$

    (cf. René L. Schilling/Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes, pp. 277) Probably, there exists a variety of other transformations and criterions, but I'm not into this topic.

  3. The approach to solve a linear SDE is rather similar to the deterministic case: First, we consider the homogenous SDE $$dZ_t = \beta Z_t \, dt + \delta Z_t \, dW_t$$ To solve this equation, we apply Itô's formula to $\log Z_t$. It's not difficult to show that the solution equals $$Z_t = Z_0 \cdot \exp \left( \left(\beta- \frac{1}{2} \delta^2 \right) \cdot t+ \delta \cdot W_t \right) \tag{1}$$ To find a solution of the non-homogenous SDE, we set $Y_t := Z_t \cdot Z_t^0$ where $\frac{1}{Z_t^0}$ is given by $(1)$. Using Itô's formula, we find an expression for $Y_t$ and since we know $Z_t^0$, this allows us to compute $Z_t$. There are a lot of calculations involved, but they are not that difficult.

Is this perhaps the "Doss-Sussmann" / "change of scale" method? See e.g. Rogers, Williams, vol. 2, Section V.28.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.