# How to solve the SDE $dX_t = aX_tdt + (b(t)-X_t^2)^{1/2}dW_t$?

I need help on solve the following SDE: $\beta > 0$, $0<\gamma<1$, $X_0 = \frac{\sqrt{2}}{2}$ $$dX_t = -(\beta + \frac{1}{2}\gamma^2)X_tdt + \gamma\sqrt{e^{-2\beta t}-X_t^2}dW_t$$

I need help to find the transform function $f(t,x)$ and let $Y_t = f(t,X_t)$ so that $dY_t$ can be solved.

Thank you very much.

• It seems to me that $f(t,x) = e^{-2 \beta t} - x^2$ should work Nov 7, 2014 at 15:00

Hint:

$$f(t,x) := \frac{1}{\gamma} \arctan \left( \frac{x}{\sqrt{e^{-2\beta t}-x^2}} \right)$$

transforms the given SDE to a (very simple) linear SDE.

Remark: The transformation is a so-called variance transform. For an SDE of the form $$dX_t = b(t,X_t) \, dt + \sigma(t,X_t) \, dW_t \tag{1}$$ the transformation is given by

$$f(t,x) := \int_0^x \frac{dy}{\sigma(t,y)}.$$

In fact, there are necessary and sufficient conditions (in terms of derivatives of $b$ and $\sigma$) which state whether an SDE of the form $(1)$ can be transformed via $Y_t := f(t,X_t)$ into a linear SDE of the form $$dY_t = \bar{b}(t) \, dt + \bar{\sigma}(t) \, dW_t.$$ (See e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 19, for more details.)

Solution: $$X_t = \sqrt{\frac{e^{-2 \beta t} \tan^2(\gamma W_t+\pi/4)}{1+\tan^2(\gamma W_t+\pi/4)}} = e^{-\beta t} \sin(\gamma W_t+\pi/4).$$

• Really nice, I didn't know it went under this name. Is the name linked to the way you can use it for increase convergence rate in numerical simulations of stochastic processes? Nov 7, 2014 at 16:34
• @JosephK I don't know. I just have added a reference; in this book, the transformation is called variance transform since it is defined in such a way that $\bar{\sigma}(t)=1$.
– saz
Nov 7, 2014 at 16:41
• Which can be simplified into $X_t = e^{- \beta t} \sin(\gamma W_t+\pi/4)$.
– Did
Nov 8, 2014 at 10:38
• @Did Thanks... :)
– saz
Nov 8, 2014 at 10:54