# Brownian motion, compact interval

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?

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).
• 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.. – Maria May 14 at 5:28
• 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 $(1/h^2)\cdot h = 1/h$ – Alex R. May 14 at 5:44