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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?

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  • $\begingroup$ I wrote these formulae by Microsoft Word. They look completely different here. How should I edit them? $\endgroup$ – Hebe Jul 26 '13 at 8:31
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    $\begingroup$ You can use $\LaTeX$ $\endgroup$ – Bob Jansen Jul 26 '13 at 8:47
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    $\begingroup$ Make sure you use the Latex notation next time please, it will save me a lot of time. $\endgroup$ – SRKX Jul 26 '13 at 9:21
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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.

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  • $\begingroup$ 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 '17 at 1:02
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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.

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