Itô's formula for a $\mathcal{C}^2$ function of two variables F reads: \begin{align} F(X_t, Y_t) &= F(X_0, Y_0) + \int_0^t \frac{\partial F}{dx}(X_s, Y_s) \, dY_s + \int_0^t \frac{\partial F}{dy}(X_s, Y_s) dX_s \\ &+ \frac{1}{2}\int_0^t\frac{\partial^2 F}{dx^2}(X_s, Y_s) d\langle Y, Y\rangle_s + \frac{1}{2}\int_0^t \frac{\partial^2 F}{dy^2}(X_s, Y_s) d\langle X, X\rangle_s \\ &+ \frac{1}{2}\int_0^t \frac{\partial^2 F}{dx \,dy}(X_s, Y_s) d\langle X, Y\rangle_s \end{align}

However, it seems that when applied to 1) $X_t = t$ and $Y_t = B_t$ (Brownian motion in 1D) , or 2) $X_t$ local martingale and $Y_t = \langle X, X \rangle_t$, the last two integrals disappear in the formulas I've seen.

Why is that so ?

  • 1
    $\begingroup$ Because the quadratic covariation $\langle f, g \rangle$ equals zero if $f$ is continuous and $g$ of bounded variation $\endgroup$ – saz Feb 27 at 13:00
  • 1
    $\begingroup$ Because $dtdB_t = o(t^{3/2})$ and $(dB_t)^2 = dt$. In the first case when, you expand the terms $d[X,X]_s$ and $d[X,Y]_s$, you get things which are $o(ds^\alpha)$ with $\alpha>1$ and so can be thrown away. $\endgroup$ – dohmatob Feb 27 at 15:28
  • $\begingroup$ Also see math.stackexchange.com/a/1351694/168758. $\endgroup$ – dohmatob Feb 27 at 15:34
  • $\begingroup$ Thank you, @dohmatob I think that using small o notation here is somewhat of an abuse, and also does not explain the result for the martingale. But it's still useful as mnemonics! $\endgroup$ – P. Camilleri Mar 1 at 5:23
  • $\begingroup$ My comment was for the first of your cases. For the second, you'll need to provide more details / refs. It's usually best to post questions which are self-contained, in terms of terminology, etc. $\endgroup$ – dohmatob Mar 1 at 10:18

Your Answer

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

Browse other questions tagged or ask your own question.