Let $W_{t}$ be a brownian motion and $$ W^{*}_{t} = \max_{s<t} W_{s} $$

Then can you please explain why we have this: $$ (W^{*}_{t} - W_{t})dW^{*}_{t} = 0 $$


Process $W_t^\ast = \max_{s < t} W_s$ is caglad (left-continuous with the right limit) piecewise constant:

Trajectory of the W* process

Therefore, intuitively speaking, $\mathrm{d} W_t^\ast$ is mostly zero. At the points of discontinuities of $W_t^\ast$, the Wiener process $W_t$ crosses $W_t^\ast$ from below, meaning that $W_t^\ast = W_t$. Now whether this makes the product $(W_t-W_t^\ast) \mathrm{d} W_t^\ast$ zero depends on details of what you mean by $\mathrm{d}W_t^\ast$, which you did not provide.

  • $\begingroup$ I think that having pointed out that dWt = 0 on {t: Wt > Wt}, your proof is completed by acknowledging that the Lebesgue measure of the set {t: W*t = Wt} is zero, and this fact follows from the infinite variation property of the brownian motion. $\endgroup$ – user48531 Nov 7 '12 at 17:23

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.