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I am looking to compute the quadratic variation of $$S_t = S_0e^{\sigma B_t}$$ where $B_t$ is Brownian Motion. Applying Itô's lemma, I having the following $$(dS_t)^2 = S_0^2\sigma^2e^{2\sigma B_t}dt$$ Now here is where I am a bit confused... how is this actually computed?

I know that the following can't be solved using traditional calc (given that $B_t$ is not differentiable) $$S_0^2\sigma^2\int_0^Te^{2\sigma B_t}dt$$

Do I apply Ito's lemma again, assuming something like $$f_{xx}(t,x) = e^{2\sigma x}$$ and assume that f is not a function of t?

...Or do I approximate with something like $$\sum_ie^{2\sigma B_{i}}(t_{i+1}-t_i)$$

Having a tough time finding any detail in the literature here - any help is appreciated.

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If we have $$S_t = S_0 e^{\sigma B_t},$$

then Itô expanding it gives \begin{align} dS &= \frac{\partial S}{\partial t} dt + \frac{\partial S}{\partial B} dB + \frac{1}{2}\frac{\partial^2 S}{\partial B^2} (dB)^2 \\ &= 0 \cdot dt + \sigma S_0 e^{\sigma B}dB + \frac{1}{2}\sigma^2 S_0 e^{\sigma B} dt \\ &= \frac{1}{2}\sigma^2 S_t dt + \sigma S_tdB, \end{align} where we used that $(dB)^2=dt$.

Hence, we have that \begin{align} d\langle S \rangle_t := (dS_t)^2 = \sigma^2S_t^2dt, \end{align} so that the quadratic variation is given by $$\langle S \rangle_t = \sigma^2 \int_0^t S_u^2 du.$$

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  • $\begingroup$ @Pontius: I followed most of it except for the "Hence, we have that" part. Could you explain that ? Thanks. $\endgroup$
    – mark leeds
    Dec 10, 2022 at 15:44
  • $\begingroup$ @markleeds: Since dS=0.5*v^2*Sdt + vSdB, we have that (dS)^2 = ...dt^2 + ...dtdB + ...dB^2 = ...dB^2 = ...dt, since dt^2~0, dt*dB~0, and dB^2 = dt. Quadratic variation is then given by integrating (dS^2) $\endgroup$ Dec 13, 2022 at 15:33
  • $\begingroup$ Thanks Pontus. I get it now. It's appreciated. $\endgroup$
    – mark leeds
    Dec 13, 2022 at 22:07

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