If we know the dynamics of $S$, then we can estimate the value of $S$ at a time point, $t$. Here, I have a question concerning how to solve for $S_t$ by Itô because I obtained different results by different approaches.

For a geometric Brownian motion: $$dS_t=S_t μ dt+S_t σdW_t,$$ $$\frac{dS_t}{S_t} =μ dt+σdW_t,$$ and, in fact we have, $$\frac{dS_t}{S_t} =d\ln(S_t).$$ If we make $Z=d\ln(S_t)$, then, $$dZ=\frac{\partial Z}{\partial t} dt+\frac{\partial Z}{\partial S_t} dS_t+ \frac{1}{2} \frac{\partial^2 Z}{\partial S_t^2} (dS_t)^2=(μ-\frac{1}{2} σ^2 )dt+σdW_t,$$

$$Z_t= Z_0+\left(μ- \frac{σ^2}{2} \right) \int_0^tds+σ\int_0^tdW_s,$$ $$\ln(S_t )=\ln(S_0 )+(μ-\frac{1}{2} σ^2 )dt+σW_t,$$ $$S_t=S_0 \cdot e^\left((μ- \frac{1}{2} σ^2 )dt+σW_t \right).$$

However, if I use another approach, then I get the different result. Since we have $\frac{dS_t}{S_t} =d\ln(S_t)$ then, $$d\ln(S_t )=μdt+σdW_t$$ and \begin{align} \ln(S_t)&=\ln(S_0)+μ\int_0^tds+σ\int_0^t dW_s \\ &=\ln(S_0)+μt + σW_t, \end{align} $$S_t=S_0 \cdot e^{(μt+σW_t)}$$

I think both approaches are correct. But why are the results distinct?

  • $\begingroup$ I wrote these formulae by Microsoft Word. They look completely different here. How should I edit them? $\endgroup$
    – Hebe
    Jul 26, 2013 at 8:31
  • 1
    $\begingroup$ You can use $\LaTeX$ $\endgroup$
    – Bob Jansen
    Jul 26, 2013 at 8:47
  • 4
    $\begingroup$ Make sure you use the Latex notation next time please, it will save me a lot of time. $\endgroup$
    – SRKX
    Jul 26, 2013 at 9:21

2 Answers 2


The part where you say that

$$\frac{dS_t}{S_t} = d\ln(S_t)$$

is wrong, because $S$ is a stochastic variable.

This is exactly what Itô tells you with his formula that you apply right do compute your $dZ$.

The difference comes from the quadratic variation of the process $S$ which you express as $(dS)^2$. If you don't add this term when the variable are stochastic, your derivation is wrong.

  • $\begingroup$ So $\frac{d\ln S}{d S} \neq 1/S$, but why it's okay to say $\frac{\partial \ln S}{\partial S} = 1/S$? $\endgroup$
    – GuLearn
    Jul 31, 2017 at 1:02

You're right, both approaches are correct in a way (but I think you have some messiness in how you written everything out as @SRKX pointed out) ... but under different formulations of stochastic calculus. Your second answer is the solution for the Stratonovich SDE:

$$\text{d}S_t = \mu S_t \text{d}t + \sigma S_t \circ \text{d}W_t,$$

Under the Stratonovich interpretation the generic calculus chain rule applies, so you don't need a form of Itô's formula/lemma, i.e., the chain rule for Itô stochastic calculus. These chain rules are used to remove state dependence (in your case, dependence on $S_t$) from the stochastic integrals that correspond to the SDEs, allowing them to be solved.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.