# Geometric Brownian motion and its inverse

Consider a geometric Brownian motion described by the SDE:

$$dS_t = \mu S_t dt + \sigma S_t dB_t$$

where $$B_t$$ is a Brownian motion and $$\mu$$ and $$\sigma$$ are constants. We have that:

$$d(\ln S_t) = \sigma dB_t +\left(\mu - \frac{\sigma^2}{2}\right)dt$$

from which we immediately obtain:

$$S_t = S_0 \exp\left[(\sigma B_t)+\left(\mu - \frac{\sigma^2}{2}\right)t\right]$$

Consider now the process $$R_t = S_t^{-1}$$; in this case, since $$dR_t = - S_t^{-2} dS_t$$, we obtain the SDE:

$$dR_t = -\mu R_t dt - \sigma R_t dB_t$$

and by applying the same procedure, we obtain:

$$R_t = R_0 \exp\left[(- \sigma B_t)+\left(- \mu - \frac{\sigma^2}{2}\right)t\right]$$

On the other hand, from inverting immediately the $$S_t$$ solution I would obtain:

$$R_t = R_0 \exp\left[(- \sigma B_t)+\left(- \mu + \frac{\sigma^2}{2}\right)t\right]$$

Obviously there's something wrong here, but which method is correct? And where's the mistake?

The mistake is that the SDE $$dR_t = - S_t^{-2} dS_t$$ is not correct. By applying Itô's formula to $$R_t = S_t^{-1}$$ we instead get: \begin{align*} dR_t &= -S_t^{-2}dS_t + S_t^{-3}(dS_t)^2 \\ &= - \mu R_t dt - \sigma R_t dB_t + \sigma^2 R_t dt\\ &= (\sigma^2 - \mu)R_tdt - \sigma R_t dB_t \end{align*} Solving this SDE by the method you propose leads to $$R_t = R_0 \exp \left( -\sigma B_t + (-\mu + \sigma^2 /2)t \right)$$ which is what you expect.