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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.

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

1 Answer 1

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

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  • $\begingroup$ 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? $\endgroup$
    – JosephK
    Nov 7, 2014 at 16:34
  • $\begingroup$ @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 $\bar{\sigma}(t)=1$. $\endgroup$
    – saz
    Nov 7, 2014 at 16:41
  • $\begingroup$ Which can be simplified into $X_t = e^{- \beta t} \sin(\gamma W_t+\pi/4)$. $\endgroup$
    – Did
    Nov 8, 2014 at 10:38
  • $\begingroup$ @Did Thanks... :) $\endgroup$
    – saz
    Nov 8, 2014 at 10:54

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