# Brownian Motion Differential Equation

The behavior of $S(t)$ through time is modeled by $ππ(π‘) = π\,ππ‘ + π\,ππ(π‘)$ for a $π(π‘)$ οΏΌstandard Brownian motion and real value ΞΌ and Ο > 0. Now, let $π(π‘) = \frac{1}{S(t)}$ Show that $U(t)$ satisfies the following stochastic differential equation. $$ππ(π‘) = (π^2 β π)\,ππ‘ β π\,ππ(π‘)$$

I solved the DE regarding $S(t)$ and get $S=S_oe^{\mu t-\frac{π^2}{2}t+πWt}$ And then how should I get to $U(t)$. Can anyone help me to get it? Thank you!

• Can anyone help me with it? Thankyou! Aug 7, 2017 at 3:01
• Apply Itô's formula...
– saz
Aug 7, 2017 at 4:14
• @saz Hi, thank you. But can you be more specific? Aug 7, 2017 at 5:15
• Find a function $f$ such that $U_t = f(t,W_t)$. Now apply Itô's formula: $$f(t,W_t)- f(0,W_0) = \dots$$
– saz
Aug 7, 2017 at 5:17
• Your solution is the one for the geometric Brownian motion, your equation has $S(t)=S(0)+ΞΌ·t+Ο·W(t)$ as solution by simple integration if the coefficients are constant. Aug 8, 2017 at 15:40

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 $$dY_t = Y_t\left[ \left( \alpha\mu +\frac{1}{2}\alpha(\alpha-1)\sigma^2 \right)dt+\alpha\sigma\,dW_t \right]$$