Brownian Motion Differential Equation The behavior of $S(t)$ through time is modeled by $() = \, + \,()$ for a $()$ ￼standard Brownian motion and real value μ and σ > 0. Now, let $() =
\frac{1}{S(t)}$ Show that $U(t)$ satisfies the following stochastic differential equation.
$$() = (^2 − )\, − \,()$$
I solved the DE regarding $S(t)$ and get $S=S_oe^{\mu t-\frac{^2}{2}t+Wt}$
And then how should I get to $U(t)$.
Can anyone help me to get it? Thank you!
 A: It seems like there might be some typos in your question. Firstly, $S_t$ is not a standard Brownian motion since it has a non-zero "drift term" and non-unity "diffusion coefficient". Secondly, the equation:
$$ dS_t = \mu \,dt + \sigma\,dW_t $$
has solution $$ S_t=\mu t+\sigma W_t +S_0 $$
On the other hand, geometric Brownian motion (GBM) satisfies:
$$ dX_t = X_t( \mu \,dt + \sigma\,dW_t ) $$
and has solution (as you found) $$ X_t=X_0\exp\left( \left[\mu-\frac{\sigma^2}{2}\right]t+\sigma W_t \right) $$
In any case, let's apply Ito's lemma to GBM: $$ df(t,X_t)=\partial_t f(t,X_t)\,dt+\partial_xf(t,X_t)\,dX_t+\frac{1}{2}\partial_{xx}f(t,X_t)[dX_t]^2 $$
So we set $U_t= 1/X_t$ (i.e. $U_t=f(X_t)$, where $f(X_t)=X_t^{-1}$. Thus, by Ito's Lemma:
\begin{align}
dU_t &= -X_t^{-2}\,dX_t+\frac{1}{2}(2X_t^{-3})\sigma^2X_t^2\,dt\\
&= \frac{-X_t}{X_t^2}(\mu\,dt + \sigma\,dW_t)+\frac{\sigma^2}{X_t}dt\\
&= \frac{1}{X_t}(-\mu\,dt-\sigma\,dW_t+\sigma^2\,dt)\\
&= U_t([\sigma^2-\mu]dt-\sigma\,dW_t)
\end{align}
This is not the same as what you get; perhaps you are missing a $U_t$ factor?

Note: in general, if $Y_T=(X_t)^\alpha$ and $X_t$ is 1D GBM, then
$$ dY_t = Y_t\left[ \left( \alpha\mu +\frac{1}{2}\alpha(\alpha-1)\sigma^2 \right)dt+\alpha\sigma\,dW_t \right] $$
