Consider the following SDE with $X_0 = 1$, $$ dX_t = X_t\operatorname{sign}(X_t) \, dt + X_t \, dW_t, $$ where $\operatorname{sign}(x) = \mathbb{1}\{x \ge 0\}$. How am I supposed to solve this SDE?
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$\begingroup$ show it never hits zero, then the sign($X_t$) is moot. $\endgroup$– mikeCommented Feb 8, 2021 at 9:17
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$\begingroup$ @mike I did think about this, but I do not know how to show it never hits zero. $\endgroup$– Van TomCommented Feb 8, 2021 at 9:19
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6$\begingroup$ unitl it hits 0 it satisfies the equation of geometric brownian motion, and since a geometric brownian motion does not hit 0, you are done. $\endgroup$– mikeCommented Feb 8, 2021 at 12:09
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1 Answer
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Indeed, as mentioned in the comments one can return to a geometric BM. We start with the Lamperti transformation 2.1.5/2.1.6 from "The Lamperti Transform" by de Boer. First using Ito to compute for general map $\psi$.
then for this specific $\psi$ below, they turn the volatility coefficient to be one.
So here we have $Z_{t}=\psi(X_{t})=\log X_{t}$ and so it satisfies the SDE for geometric BM
$$dZ_{t}=(0+\frac{e^{Z_{t}}1\{e^{Z_{t}}\geq 0\}}{e^{Z_{t}}}-\frac{1}{2})+dW_{t}=\frac{1}{2}dt+dW_{t}.$$
answered Jan 14, 2023 at 8:23