(I offer 100 bounty because I really would like to have a constructive answer to this question)

  • I often see that if $W_t$ is a Brownian motion, then $dW_t\sim (dt)^{1/2}$. Can it come from the fact that Brownian motion is $1/2-$holder continuous and not better (i.e. $$\sup_{t,s\in [0,1], t\neq s}\frac{|B_t-B_s|}{|t-s|^\alpha }=\infty\quad a.s.$$ for all $\alpha \in (1/2,1]$.) ? Indeed, we have that $$|dW_t|=|W_{t+h}-dW_t|\leq C|h|^{1/2}=C(dt)^{1/2},$$ so $dW_t=\mathcal O(dt^{1/2}),$ but $dW_t\neq \mathcal O(dt^{\alpha })$ for all $1/2<\alpha<1$. Does this make sense ?

  • If yes, does it imply that the quadric variation is bounded, or quadric bounded variation implies $1/2-$holder continuity and not better ?


If we have a nice function $f$, then we can write its Taylor series, $$ f(x+h) = f(x) + f'(x)h + \frac12f''(x)h^2 + \frac1{3!}f'''(x)h^3 + \cdots. $$ Let $\Delta W_t=W_{t+\Delta t}-W_t$. Then, taking $x=W_t$ and $h=\Delta W_t$, we get $$ f(W_{t+\Delta t}) = f(W_t) + f'(W_t)\Delta W_t + \frac12 f''(W_t)(\Delta W_t)^2 + \frac1{3!}f'''(W_t)(\Delta W_t)^3 + \cdots. $$ Now fix $T>0$. Let $\Delta t=T/n$ and let $t_j=j\Delta t$. Then \begin{align} f(W_T) &= \sum_{j=0}^{n-1} (f(W_{t_{j+1}}) - f(W_{t_j}))\\ &= \sum_{j=0}^{n-1} (f(W_{t_j + \Delta t}) - f(W_{t_j}))\\ &= \sum_{j=0}^{n-1} f'(W_{t_j})\Delta W_{t_j} + \frac12\sum_{j=0}^{n-1} f''(W_{t_j})(\Delta W_{t_j})^2 + \frac1{3!}\sum_{j=0}^{n-1} f'''(W_{t_j})(\Delta W_{t_j})^3 + \cdots. \end{align} The first sum converges in an appropriate sense to an Ito integral: $$ \sum_{j=0}^{n-1} f'(W_{t_j})\Delta W_{t_j} \to \int_0^T f'(W_t)\,dW_t. $$ The second sum converges in an appropriate sense to an ordinary integral: $$ \frac12\sum_{j=0}^{n-1} f''(W_{t_j})(\Delta W_{t_j})^2 \to \int_0^T f''(W_t)\,dt. $$ This convergence is fundamentally connected to the fact that the quadratic variation of Brownian motion on the interval $[0,t]$ is $t$. The notation $(dW_t)^2 \sim dt$, or $dW_t \sim (dt)^{1/2}$, is a shorthand heuristic for this convergence and the rules and results that follows from it.

Because Brownian motion has a finite quadratic variation, it's $p$-variation for $p>2$ is zero. Because of this, all of the sums with $(\Delta W_{t_j})^n$, where $n>2$, tend to zero. This is how we derive Ito's rule.

Regarding Holder continuity, there is a fundamental connection between it and $p$-variation. (See https://en.wikipedia.org/wiki/P-variation#Link_with_Hölder_norm.) However, one must be careful when using this. The $p$-variation referenced there is the one used in analysis for deterministic functions. We do not define the quadratic variation of a stochastic process by simply applying the deterministic definition of the $2$-variation to each sample path. (See https://en.wikipedia.org/wiki/Quadratic_variation.) Rather, there is a subtle difference involving the choice of partitions and the type of convergence. Because of this subtle difference, you can have counterintuitive results. For example, Brownian motion has a finite quadratic variation, but it's sample paths have infinite $2$-variation.

  • $\begingroup$ I'm confuse about something... can the fact that $dW_t\sim (dt)^{1/2}$ be explained using the $1/2-$holder continuity and not better ? Because for me, my argument makes sense. What do you think ? (then I will accept your answer). $\endgroup$ – user657324 May 4 at 14:54
  • $\begingroup$ It's not 100% clear what you're asking, since $dW_t\sim(dt)^{1/2}$ is not a mathematical fact that requires (or has) a proof or explanation. Rather, it is a heuristic bit of notation that refers to the convergence I indicated above. That convergence has a proof which uses the quadratic variation (not Holder continuity and not 2-variation) of Brownian motion. So although Holder continuity is related to the notation, I would not say it is the basis or central idea of it. $\endgroup$ – Jason Swanson May 4 at 15:10
  • $\begingroup$ Indeed, but don't you think so that the fact that $|B_{t+h}-B_t|\leq C|h|^{1/2}$ a.s. could make sense to the fact that if $|\Delta W_t|\leq C|\Delta t|^{1/2}$, and thus if $\Delta t$ is small enough, then why not $dW_t\sim (dt )^{1/2}$ (even if it's in a formal way). $\endgroup$ – user657324 May 4 at 15:44
  • $\begingroup$ A more notationally consistent way to translate $|\Delta W_t|\le C|\Delta t|^{1/2}$ into a heuristic involving differentials would be $dW_t=O((dt)^{1/2})$. In the context of asymptotic behavior, "$\sim$" usually means the ratio tends to 1. But this is not true (see law of iterated logarithm). $\endgroup$ – Jason Swanson May 4 at 15:55
  • $\begingroup$ Good point ! Thanks a lot :-) $\endgroup$ – user657324 May 4 at 16:14

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.