# Name of the formula transforming general SDE to linear

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 $$d(x) := \int_0^x \frac{1}{\sigma(y)} \, dy \qquad \qquad \delta = - \frac{\frac{d}{dx}(\kappa'(x) \cdot \sigma(x))}{\kappa'(x)}$$

• @saz What is $\gamma$? – Kerry Nov 7 '15 at 18:37
• @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. – 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.
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 $$0 = \frac{\partial}{\partial x} \left( \sigma(t,x) \cdot \left( \frac{\sigma_t(t,x)}{\sigma^2(t,x)} - \frac{\partial}{\partial x} \frac{b(t,x)}{\sigma(t,x)} + \frac{1}{2} \sigma_{xx}(t,x) \right) \right)$$
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.