# 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?

• (a) Apply Ito's lemma on $S(t)$ and $f: x \mapsto x^2$; (b) Once you have the expression of $X(t)$, in that form, you can compute the quadratic variation term by using only the stochastic part. – byouness Mar 11 at 8:40
• Please don't destructively edit your questions. – Bob Jansen Jul 6 at 5:45

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 $$X_t=S_t^2$$.