Quadratic Variation of an Exponential Brownian Price Process
Summary
The document explains how to obtain the quadratic variation of a process defined as an exponential function of Brownian motion. Applying Itô’s lemma gives the price process a drift term and a Brownian shock term. In the quadratic variation, the drift contributes no instantaneous variation, while the squared shock term yields a rate proportional to the squared price and volatility squared.
Integrating that rate over time gives quadratic variation as a time integral of the squared process. The answer does not evaluate this integral into a closed-form random variable; it identifies the correct accumulated quantity, which can be approximated by sums over time partitions. This distinction addresses the question’s concern that Brownian paths are not differentiable: stochastic calculus defines the variation through the squared increments rather than ordinary differentiation.
Key ideas
- Itô’s lemma gives the exponential Brownian process both drift and diffusion terms.
- Quadratic variation accumulates the squared diffusion coefficient over time.
- The resulting variation is the integral of volatility squared times the squared price.
- The integral need not be evaluated by ordinary calculus to define quadratic variation.
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# Integral of Function of Brownian Motion w.r.t Time (Context: Computing Quadratic Variation)
# Integral of Function of Brownian Motion w.r.t Time (Context: Computing Quadratic Variation)
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 by Pontus Hultkrantz (score 1)
https://quant.stackexchange.com/a/71584
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 $$\langle S \rangle_t = \sigma^2 \int_0^t S_u^2 du.$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.