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?

  • $\begingroup$ (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. $\endgroup$ – byouness Mar 11 at 8:40
  • $\begingroup$ Please don't destructively edit your questions. $\endgroup$ – 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 $X_t=S_t^2$.

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