I recently read this from a book:

The canonical SDE in financial math, the geometric Brownian motion, ${{d{S_t}} \over {{S_t}}} = \mu dt + \sigma d{W_t}$ has solution $${S_t} = {S_0}{e^{(\mu - {1 \over 2}{\sigma ^2})t + \sigma {W_t}}}$$ which is always positive. Again verify with Ito's lemma. Also try Ito's lemma on log $(S_{t})$

and was curious about what would it look like if you tried Itos lemma on log (St)?


1 Answer 1


We have that \begin{equation} dS_t=\mu S_t dt + \sigma S_t dW_t \end{equation}

Now apply Itô $$ d\log S_t= \frac{\partial\log S_t}{\partial t} +\frac{\partial \log S_t}{\partial S_t} dS_t + \frac12 \frac{\partial^2 \log S_t}{\partial S_t^2}d\langle S_t,S_t\rangle $$ We have that $\frac{\partial\log S_t}{\partial t}=0$ because the function $f(S_t,t)=\log S_t$ doesn't directly depend on $t$. Moreover, since $\frac{d}{dx} \log x=\frac{1}{x}$ and $\frac{d^2}{dx^2} \log x=-\frac{1}{x^2}$, we obtain the following formula: $$ d\log S_t= \frac{1}{S_t} dS_t -\frac12 \frac{1}{S_t^2}d\langle S_t,S_t\rangle $$ Just use the fact that the quadratic variation of $S_t$ is simply $\langle S_t,S_t\rangle =\sigma^2 S_t^2$ and the first equation to obtain: $$d \log S_t = \left(\mu -\frac{\sigma^2}{2}\right) dt +\sigma dW_t $$ Integrate and obtain $$\log S_t = \log S_0 + (\mu-\frac12 \sigma^2)t +\sigma W_t$$ If you apply Itô's lemma you get back to the previous point. If you use the exponential you finally have your result: $$S_t=S_0 e^{(\mu-\frac12 \sigma^2)t +\sigma W_t}$$ Tadaaa

  • $\begingroup$ thank you ! basically he wants to show that you get the same solution even in the log st case? $\endgroup$
    – k b
    Commented Mar 30, 2017 at 13:29
  • 1
    $\begingroup$ No basically the author wants you to (1) apply Itô's lemma to the solution $S_t = S_0 e^{(\mu-\frac{1}{2}\sigma^2)t + \sigma W_t} $ to find that it indeed verifies the SDE $dS_t/S_t = \mu dt + \sigma dW_t$; (2) by applying Itô's lemma to $f(t,S_t)=\ln(S_t)$, understand where the solution $S_t= S_0 e^{(\mu-\frac{1}{2}\sigma^2)t + \sigma W_t}$ actually comes from (see NSZ's nice answer). $\endgroup$
    – Quantuple
    Commented Mar 30, 2017 at 14:05

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