My question is about the notation used in these stochastic differential equations.
Let $X_t$ be a stochastic process satisfying:
$\displaystyle X_t = X_0 + \int_0^t \mu(s,\omega) \, \mathrm d s + \int_0^t \nu(s,\omega) \, \mathrm d B_s$
Shorthand:
$\mathrm dX_t = \mu_t \, \mathrm dt + \nu_t \mathrm d B_t$
where the last integral is a Brownian motion integral, and
where $\mu(t,\omega), \nu(t, \omega)$ are $\mathcal F_t$ adapted $L^2$ functions. (In class we so far only did the case where $\mu$ and $\nu$ are deterministic).
Let $f(t,x)$ be a twice differentiable deterministic function.
The usual presentation of Ito's lemma is:
$\displaystyle \mathrm df(t,X_t) = \left({\frac{\partial f}{\partial t} + \mu_t \frac{\partial f}{\partial x} + \frac 1 2 \frac{\partial^2 f}{\partial x^2}\nu_t^2}\right)\mathrm dt + \frac{\partial f}{\partial x}\nu_t\, \mathrm dB_t$
The professor offered us this shorthand:
$\displaystyle \mathrm df(t,X_t) = \frac{\partial f}{\partial t}\, \mathrm dt + \frac{\partial f}{\partial x}\, \mathrm dX_t + \frac 1 2 \frac{\partial^2 f}{\partial^2 x} \, (\mathrm d X_t)^2 $
He explained that the notation $(\mathrm d X_t)^2$ is to be interpreted as the taking the quadratic variation whenever the algebra would suggest you multiply the differentials. For example, $\mathrm d B_t \mathrm d B_t =\mathrm d \langle B_t, B_t \rangle_T = \mathrm d (T) \, \text{a.s.} = \mathrm dt$ (this last step has its own answer on s.e.)
Why is the formal multiplication of (stochastic) differentials interpreted as the quadratic variation?
Best Answer
Well, to my modest opinion, we only have the slightly boring answer: Because it works. Via the ito isommetry one can show, that for any bounded, predictable $f^1, f^2$ and semimartingales $X$, $Y$ one has:
$\langle \int_0^\cdot f^1_s dX_s, \int_0^\cdot f^2_s dY_s \rangle_t = \int_0^t f^1_s f^2_s d\langle X,Y\rangle_s$,
which is basically the justification of doing it. Furthermore, one can even make it rigorous with some algebraic structures.