# What is the SDE of this equation? [closed]

I am new and struggling to understand how to solve this using Ito lemma.

Can someone please explain it to me:

$$dS_t=-\frac{1}{2}\sigma^2 S_t dW_t$$

what is the solution with explanation please

## 1 Answer

Actually this is just the Black-Scholes SDE with zero drift and $$-\frac{1}{2}\sigma^2$$ volatility. If you plug that into the well known solution, you get $$S_t=S_0e^{\frac{1}{8}\sigma^4t-\frac{1}{2}\sigma^2 W_t}$$ but let's calculate it with Ito's formula.

Choose $$f(x)=\log(x)$$, then we have $$f'(x)=\frac{1}{x}$$ and $$f''(x)=-\frac{1}{x^2}$$. Inserting in Ito's formula yields $$d\log(S_t)=\frac{1}{S_t}dS_t+\frac{1}{2}\left(-\frac{1}{S_t^2}\right)d\langle S\rangle_t \\ =-\frac{1}{S_t}\frac{1}{2}\sigma^2S_tdW_t+\frac{1}{2}\frac{1}{S_t^2}\frac{1}{4}\sigma^4S_t^2dt \\ = \frac{1}{8}\sigma^4dt-\frac{1}{2}\sigma^2dW_t$$ or equivalently $$\log(S_t)=\log(S_0)+\frac{1}{8}\sigma^4 t-\frac{1}{2}\sigma^2W_t \\ \Leftrightarrow S_t=S_0\exp\left(\frac{1}{8}\sigma^4t-\frac{1}{2}\sigma^2W_t\right)$$

• Why is it not (-1/8sigma^4t...) – InfoLearner Jan 21 '20 at 12:26
• @InfoLearner you mean why it is the exponential of that? Because you have to get rid of the logarithm that we applied to use Ito's formula – Leguan3000 Jan 23 '20 at 18:28