4
$\begingroup$

I'm looking to find out the stochastic differential equation satisfied by the quadratic variation of

  1. Geometric Brownian Motion,
  2. Diffusion Process.

For example, for a diffusion process that satisfies $dX_t = \mu(t, X_t) dt + \sigma (t, X_t) dW_t$, I've heard that $d\langle X\rangle_t = \sigma^2dt$. But I've tried looking everywhere for a proof or verification of this and not found anything. I attempted it myself but no luck so far; I vaguely think Ito's lemma may be needed.

I already know that the QV of standard Brownian motion is simply the time $t$.

I'd appreciate if anyone could show me how the equation for QV of Diffusion and GBM comes about, or point out a reference to me that deals with this.

Thanks in advance!

$\endgroup$
5
$\begingroup$

Assume that $X$ solves $$ dX_t = \mu(t,X_t) dt + \sigma(t, X_t) dW_t. $$ In integral form, this means $$ X_t = X_0 + \int_0^t \mu(s,X_s) ds + \int_0^t \sigma(s,X_s) dW_s. $$ Now, the quadratic variation of a stochastic integral process $H\cdot Y$, where $(H\cdot Y)_t = \int_0^t H_s dY_s$, is $$ [H\cdot Y]_t = \int_0^t H_s^2 d[Y]_s. $$ You can find this in a special case as Proposition 3.2.17 of Karatzas and Shreve's "Brownian Motion and Stochastic Calculus", or in a general version for continuous semimartingales in Section IV.31 of Rogers and William's "Diffusions, Markov processes and Martingales". Using this, we obtain (with $\mu_t = \mu(t,X_t)$ and $\sigma_t = \sigma(t, X_t)$ as convenient shorthands, and $A_t = t$) $$ [X]_t = [\mu \cdot A]_t + [\sigma\cdot W]_t = \int_0^t \mu_s^2 d[A]_s + \int_0^t \sigma_s^2 d[W]_s \\ = \int_0^t \sigma(s,X_s)^2 ds, $$ where we have used $[A]_t = 0$, since all continuous processes of finite variation have zero quadratic variation, and $[W]_t = t$. As for the geometric Brownian motion, we have that $X$ is a GBM if it satisfies $$ dX_t = \mu X_t dt + \sigma X_t dW_t, $$ which yields $$ [X]_t = \int_0^t \sigma^2 X_s^2 ds. $$ In particular, $[X]$ isn't immediately seen to satisfy any SDE or ODE (as $[X]$ does not figure in the right-hand side of the above). The above is simply a representation of $[X]$ in terms of $X$, but I doubt that there in general exists and SDE or ODE satisfied by $[X]$.

$\endgroup$
  • 1
    $\begingroup$ Ah... that theorem for $H.Y$ you used seems to be the key step. I will look it up in the references you mentioned. Thanks! $\endgroup$ – ul15524 Aug 10 '14 at 15:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.