Why do we expect that Brownian motion remains in the ball $B(0,\delta )$ provides $T\ll \delta ^2$?

Question

Let $(B_t)$ a Brownian standard motion and $$\mathcal B(0,\delta )=\{\varphi \in \mathcal C_0[0,T]\mid \|\varphi \|_\infty \leq \delta \},$$ where $\mathcal C_0[0,T]$ is the set of continuous function on $f:[0,T]\to \mathbb R$ s.t. $f(0)=0$. It's written in my lecture that if $T\ll \delta ^2$ we expect that Brownian motion remains in $\mathcal B(0,\delta )$ but I don't really understand the reason. I know that $B_t\sim \mathcal N(0,t)$, so the standard deviation of $B_t$ is $\sqrt t$, but I never really understand what it really mean. Does it mean that $B_T\in [-\sqrt{T},\sqrt{T}]$ and thus if $T\ll \delta ^2$, then $[-\sqrt T,\sqrt T]\subset \mathcal B(0,\delta )$ ? But I don't see any result that says that if $\sigma $ is the standart deviation of a r.v. $Y$ then $\mu-\sigma \leq Y\leq \mu+\sigma $ where $\mu$ is the means. Any idea ?

Answer 1

To hopefully improve intuition:

• We are dealing with a random variable, hence we never know exact what the outcome will be (it's actually a measurable function, per se). The standard deviation is the square root of the expected quadratic variance (or simply, variance). The increments of a Brownian motion are normally distributed to be precise, but remember that it is continuous.

Maybe your professor wants to tell you about "stopping times" without going too much into detail.

But sure, you can use a stop time T (stopping the Brownian motion at a desired point) to guarantee that the process is inside the ball you defined, i.e. stopping it when the supremum hits $\delta$. Then $B_T$ (the stopped process) is a continuous function, with $B_0 = 0$, inside the ball.

(The Brownian motion can be dangerous, e.g. having infinite local variation, everywhere. )