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Assume we have an SDE $$dX_t=\mu(X_t)dt + \sigma(X_t)dW_t$$ where $\sigma>0$ and $W_t$ is a Wiener process. Is there a transformation $y(X_t)$ that will make the dynamics of the transformed process $Y_t=y(X_t)$ have constant volatility?

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    $\begingroup$ The (almost trivial) case is of course $\mu(X)=\mu\times X$ and $\sigma(X)=\sigma\times X$. Then,$y=\ln(X)$ yields constant vol $\endgroup$ Commented Mar 31, 2022 at 6:16

2 Answers 2

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Yes it is called the Lamperti transform. This document, in particular Theorem 2, page 7, describes what the Lamperti transform is.

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    $\begingroup$ Nice, I did not know that this topic has been studied - thanks! $\endgroup$ Commented Mar 31, 2022 at 8:51
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    $\begingroup$ Thanks, had not heard this term before. $\endgroup$
    – fes
    Commented Mar 31, 2022 at 14:49
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Consider a function $f(X_t)$. Ito's lemma gives:

$$df(X_t)=\text{time terms}+f'(X_t)\sigma(X_t)dW_t$$

Now any $f$ satisfying:

$$f'(X_t)\sigma(X_t)=\text{constant}$$

gives a constant volatility for $f(X_t)$. Solving $f$ requires specifying $\sigma(X_t)$. For example, and as pointed out by Kermittfrog in the comments, when $\sigma(X_t)=\sigma X_t$, you can set $f(X_t)=\log(X_t)$.

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