📂Stochastic Differential Equations

Lambert Transformation

Definition ¹

$$d X_{t} = f \left( t , X_{t} \right) dt + g \left( X_{t} \right) d W_{t}$$ Let’s assume that the diffusion $g$ is dependent only on $X_{t}$ and independent of time $t$, given the stochastic differential equation (SDE) as shown above. The transformation $F : X_{t} \mapsto Y_{t}$ is called the Lamperti Transformation. $$Y_{t} := F \left( X_{t} \right) = \left. \int {{ 1 } \over { g (u) }} du \right|_{u = X_{t}}$$ The obtained $\left\{ Y_{t} \right\}$ is the solution of the transformed SDE with a unit diffusion as follows. $$d Y_{t} = \left[ {{ f \left( t, X_{t} \right) } \over { g \left( X_{t} \right) }} - {{ 1 } \over { 2 }} {{ \partial g \left( X_{t} \right) } \over { \partial x }} \right] dt + d W_{t}$$

Proof

It can be verified simply using the Itô’s lemma.

Itô’s lemma: Let’s assume the Itô process $\left\{ X_{t} \right\}_{t \ge 0}$ is given. $$d X_{t} = u dt + v d W_{t}$$ If we consider a function $V \left( t, X_{t} \right) = V \in C^{2} \left( [0,\infty) \times \mathbb{R} \right)$ and set $Y_{t} := V \left( t, X_{t} \right)$, then $\left\{ Y_{t} \right\}$ is also an Itô process, and the following holds. $$\begin{align*} d Y_{t} =& V_{t} dt + V_{x} d X_{t} + {{ 1 } \over { 2 }} V_{xx} \left( d X_{t} \right)^{2} \\ =& \left( V_{t} + V_{x} u + {{ 1 } \over { 2 }} V_{xx} v^{2} \right) dt + V_{x} v d W_{t} \end{align*}$$

Explanation

The Lamperti transformation pushes complex nonlinear terms into the drift term and fixes the diffusion term to $1$ in the original Itô process.

Example

$$d X_{t} = \mu X_{t} dt + \sigma X_{t} dt$$ Consider a geometric Brownian motion. Since $f(x) = \mu x$ and $g(x) = \sigma x$, the Lamperti transformation is $$\begin{align*} d Y_{t} =& {{ f \left( t, X_{t} \right) } \over { g \left( X_{t} \right) }} - {{ 1 } \over { 2 }} {{ \partial g\left( X_{t} \right) } \over { \partial x }} dt + d W_{t} \\ =& {{ \mu X_{t} } \over { \sigma X_{t} }} - {{ 1 } \over { 2 }} \sigma dt + d W_{t} \\ =& \left( {{ \mu } \over { \sigma }} - {{ 1 } \over { 2 }} \sigma \right) dt + d W_{t} \end{align*}$$ and its solution $Y_{t}$ is as follows. $$\begin{align*} Y_{t} =& \left. \int {{ 1 } \over { g (u) }} du \right|_{u = X_{t}} \\ =& \left. \int {{ 1 } \over { \sigma u }} du \right|_{u = X_{t}} \\ =& \left. {{ 1 } \over { \sigma }} \log u \right|_{u = X_{t}} \\ =& {{ \log X_{t} } \over { \sigma }} \end{align*}$$