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Let $(W_t)_{t\geq0}$ denote a standard Brownian motion and $I=\left[a,b\right]$ a compact interval. Show that $P\left[\frac {W_{t+h}-W_t} {h} \in I\right] \rightarrow 0$ as $h\rightarrow 0$. What does this precisely mean for the differentiability of the Brownian paths?

My idea of proof is rather intuitive. The expression $P\left[\frac {W_{t+h}-W_t} {h} \in I\right] \rightarrow 0$ as $h\rightarrow 0$ reminds me of differentiability od real function of one variable. I see that for $h=0$ the fraction is $\frac {0} {0}$, which means the limit is undefined and therefore cannot belong to $I$, as it is compact. However I think it is not sufficient. Could you please help me?

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The fraction is distributed like $\mathcal{N}(0,1/h)$, after invoking the definition of Brownian motion. As $h$ gets small, this distribution gets wider-and-wider. Meaning that any finite interval will have less-and-less probability (formally justify this by writing down the normal density and showing it goes to zero everywhere).

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  • $\begingroup$ Thank you. May I ask you how you got $1/h$? I know that $W_{t+h}-W_t \sim N(0,h)$ from the definition.. $\endgroup$ – Maria May 14 at 5:28
  • $\begingroup$ It's $(1/h) \times N(0,h)$, since you're also dividing by $h$, which you can work out is distributed like $N(0,1/h)$. A nice way of remembering this is that $\mbox{Var}(aX) = a^2\mbox{Var}(X)$ which in this case gives $(1/h^2)\cdot h = 1/h$ $\endgroup$ – Alex R. May 14 at 5:44
  • $\begingroup$ And how about the second part of the question?What does this precisely mean for the differentiability of the Brownian paths? I know BM is not differentiable in general, so is this a special case? $\endgroup$ – Maria May 14 at 9:30
  • $\begingroup$ If you take the limit inside the probability, this implies Brownian motion is almost surely nondifferentiable, as the limiting distribution is 0 everywhere and hence not a probability distribution. $\endgroup$ – Alex R. May 14 at 23:35

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