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Let $\mu$ and $\sigma$ be constants and consider the SDE $dY(t)=\mu Y(t)dt+\sigma Y(t)dW(t)$ with $W(t)$ Brownian motion and $Y(0)=y_{0}$. Using the solution to the SDE, $Y(t)=y_{0}\exp[(\mu-\frac{\sigma^{2}}{2})t+\sigma dW(t)]$, I computed the expression $dY^{-1}(t)$ as following:

Define $S(t)=Y^{-1}(t)=y^{-1}_{0}\exp[(-\mu+\frac{\sigma^{2}}{2})t-\sigma dW(t)]$

If we rewrite this in the standard form we find:

$S(t)=s_{0}\exp[(\mu_{2}-\frac{\sigma^{2}_{2}}{2})t+\sigma_{2}dW(t)]$ with $\sigma_{2}=-\sigma$, $\mu_{2}=\sigma^{2}-\mu$ and $s_{0}=y^{-1}_{0}$ (assuming $y_{0}\neq0$).

which would imply that if we 'reverse' the SDE problem we find

$dY^{-1}(t)=dS(t)=\mu_{2}S(t)dt+\sigma_{2}S(t)dW(t)=(\sigma^{2}-\mu)Y^{-1}(t)dt+(-\sigma)Y^{-1}(t)dW(t)$

Is this way of computing $dY^{-1}(t)$ correct?

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Given that $\mu$, $\sigma^2$, were general, yes it's correct. You can also check using Ito.

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