# Computing $dY^{-1}(t)$ using the SDE $dY(t)=\mu Y(t)dt+\sigma Y(t)dW(t)$

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?

## 1 Answer

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