# Is the fact that $dW_t\sim (dt)^{1/2}$ come from the $1/2-$holder property of Brownian motion?

(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 ?

## 1 Answer

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.

• 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). – user657324 May 4 at 14:54
• 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. – Jason Swanson May 4 at 15:10
• 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). – user657324 May 4 at 15:44
• 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). – Jason Swanson May 4 at 15:55
• Good point ! Thanks a lot :-) – user657324 May 4 at 16:14