Finding the SDE Let $X_t = e^{(\mu-\frac{\sigma^2}{2})t + \sigma B_t}$, where $B_t$ is a standard Brownian motion. How do I find the SDE satisfied by $X_t^{-1}?$ I know I must use Ito's formula but not sure how to apply that. Any help would be appreciated.
 A: First, let $Y_t := \ln(X_t^{-1}) = -(\mu-\frac{\sigma^2}{2})t - \sigma B_t$ (which is well defined). We see that $Y_t$ is a solution to the SDE
$$ dY_t = -(\mu-\frac{\sigma^2}{2})dt - \sigma dB_t \tag1 $$
Which means that $Y_t$ is a regular drift-diffusion process, so applying Itô's lemma to $g(t,Y_t) = e^{Y_t} = f(Y_t)$ gives
$$df(Y_t) = f'(Y_t)dY_t + \frac{1}2f''(Y_t)d\langle Y\rangle_t \tag2$$
And the quadratic variation of $Y$ is given by $d\langle Y\rangle_t =(dY_t)^2 =  \sigma^2dt$, so replacing in (2), we get
$$\begin{align} d(X^{-1}_t) = df(Y_t) &= f'(Y_t)dY_t + \frac{1}2f''(Y_t)\sigma^2dt \\ 
&= f(Y_t)\cdot(dY_t + \frac{\sigma^2}2dt) \\
&= f(Y_t)\cdot\left(-(\mu-\frac{\sigma^2}{2})dt - \sigma dB_t + \frac{\sigma^2}2dt\right) \\
&= f(Y_t)\cdot\left(-(\mu-\sigma^2)dt - \sigma dB_t\right)\\
&= -X_t^{-1}\cdot \left(\sigma dB_t + (\mu-\sigma^2)dt\right)\end{align} $$
Finally, $X^{-1}_t$ is solution to the SDE
$$d(X^{-1}_t) = X_t^{-1}\cdot \left(-\sigma dB_t + (\sigma^2 - \mu)dt\right)\qquad \square$$
Another way to prove it could have been to show that $X_t$ is an Itô diffusion process and applying Itô's lemma to $X_t^{-1} := {1 \over X_t} = f(X_t)$ directly (I haven't tried).
