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


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 $$ \log(S_t)=\log(S_0)+\frac{1}{8}\sigma^4 t-\frac{1}{2}\sigma^2W_t \\ \Leftrightarrow S_t=S_0\exp\left(\frac{1}{8}\sigma^4t-\frac{1}{2}\sigma^2W_t\right) $$

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    $\begingroup$ Why is it not (-1/8sigma^4t...) $\endgroup$ – InfoLearner Jan 21 '20 at 12:26
  • $\begingroup$ @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 $\endgroup$ – Leguan3000 Jan 23 '20 at 18:28

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