Applying Itô’s Lemma and Quadratic Variation to a Squared Price Process
Summary
The document considers the square of a geometric Brownian motion and asks how to derive its stochastic differential and quadratic variation. Applying Itô’s lemma to the function that squares the price adds a drift adjustment from the second derivative, while scaling the diffusion term by the derivative. The resulting squared process is itself a geometric Brownian motion, with adjusted drift and volatility coefficients.
For an Itô process, quadratic variation accumulates the square of the diffusion coefficient over time. Applying that rule to the squared price yields an integral involving the fourth power of the original price and the squared volatility parameter. The answer provides the derivation and notes that the expected quadratic variation could be computed using Fubini’s theorem, but does not carry out that calculation. The result assumes the stated geometric Brownian motion model.
Key ideas
- Itô’s lemma derives the differential of a nonlinear function of a stochastic process.
- Squaring a geometric Brownian motion changes its drift and scales its diffusion coefficient.
- Quadratic variation is the time integral of the squared diffusion coefficient.
- The quadratic variation of the squared price depends on the fourth power of the original price.
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# Finding Differential and Quadratic Variation Squared Process
# Finding Differential and Quadratic Variation Squared Process
A question based from Springer's Stochastic Calculus for Finance II book - I've tried working this out, but keep ending up in circles.
Let $S(t)$ be given by the usual formula for an asset price process with positive constants $\alpha$ and $\sigma$:
$$S(t)=S(0)\exp\left(\sigma W(t)+\left(\alpha-\frac{1}{2}\sigma^2\right)t\right).$$
(a) If $X(t)=S(t)^2$, how can I calculate the differential $dX(t)$ in a way such that it is of the form = $...dW(t)+...dt$?
(b) How can I calculate the quadratic variation $[X, X](t)$? Is there a general rule of thumb when moving from the differential to quadratic variation?
## Answer by Kevin (score 3)
https://quant.stackexchange.com/a/51574
As @byouness pointed out, the answer to question (a) is Itô's Lemma. You know that $\mathrm{d}S_t=\alpha S_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t$, i.e. $(S_t)$ is a geometric Brownian motion. Let $f(x)=x^2$ with $f_x=2x$ and $f_{xx}=2$. Then, $X_t=f(S_t)=S_t^2$ and
\begin{align*} \mathrm{d}X_t &=\left(\alpha S_t f_x+\frac{1}{2}\sigma^2 S_t^2 f_{xx}\right)\mathrm{d}t+\sigma S_t f_x \mathrm{d}W_t \\ &= \left(2\alpha S_t^2 +\frac{1}{2}\sigma^2 S_t^2 2\right)\mathrm{d}t+\sigma S_t 2S_t\mathrm{d}W_t \\ &= 2\left(\alpha +\frac{1}{2}\sigma^2\right)X_t\mathrm{d}t+2\sigma X_t\mathrm{d}W_t, \end{align*} i.e. $X_t=S_t^2$ is again a geometric Brownian motion. In fact, any power of a geometric Brownian motion, $S_t^n$, is again a geoemtric Brownian motion.
Regarding part (b), recall that for any Itô process $\mathrm{d}X_t=\mu(t,X_t)\mathrm{d}t+\sigma(t,X_t)\mathrm{d}W_t$, the quadratic variation of $X_t$ is given by $$[X,X]_t=\int_0^t\sigma(s,X_s)^2\mathrm{d}s,$$ i.e. $\mathrm{d}[X,X]_t=\sigma(t,X_t)^2\mathrm{d}t$. In our case, $X_t$ is a geometric Brownian motion with $\sigma(t,X_t)=2\sigma X_t$. Thus, \begin{align*} [X,X]_t=4\sigma^2\int_0^t S_s^4\mathrm{d}s. \end{align*} Using Fubini's theorem, you could at least compute the expected quadratic variation of $X_t=S_t^2$.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.