Quadratic Variation of Diffusion Process and Geometric Brownian Motion I'm looking to find out the stochastic differential equation satisfied by the quadratic variation of 


*

*Geometric Brownian Motion,

*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!
 A: 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 $[X]$.
