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Given a diffusion $dX_t = \mu(X_t)dt + \sigma(X_t)dW_t$, if one applies the transform

$f(x) = \int_a^x \frac{1}{\sigma(u)}du$, one can use Ito's lemma to show that

$df(X_t) = \left( \frac{\mu(X_t)}{\sigma(X_t)} - \frac{\sigma'(X_t)}{2} \right)dt + dW_t$.

Is there a way to explicitly invert this transform in the general case, assuming $\sigma$ is sufficiently nice?

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1 Answer 1

If you take $\mu(u)=\sigma(u)=\alpha+u^2$ then the image diffusion is the same for every $\alpha$.

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