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!


1 Answer 1


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 $[X]$.

  • 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, 2014 at 15:28

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