Integral of Function of Brownian Motion w.r.t Time (Context: Computing Quadratic Variation)

Question

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.

Answer 1

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 $$\langle S \rangle_t = \sigma^2 \int_0^t S_u^2 du.$$