Is there a concrete meaning of Brownian motion $W_t$ has variance $\sigma ^2t$?

Question

Let $(W_t)$ a Brownian motion. In particular, $W_t=\mathcal N(0,\sigma ^2t)$. So the fact that $\mathbb E[W_t]=0$ mean that the process $(W_t)_t$ gravitate around $0$. But how could w interpret $Var(W_t)=\sigma ^2t$ ? Could it be that $-\sigma \sqrt t\leq W_t\leq \sigma \sqrt t$ with very high probability ? In particular, if $t\leq T$, we can expect that $(W_t)_{t\leq T}$ won't exit the interval $[-\sigma \sqrt T,\sigma \sqrt T]$ ? Does this interpretation works or it's a bit more complicated ?

Answer 1

The fact that $-\sigma \sqrt t\leq W_t\leq \sigma \sqrt t$ with high probability is not completely true. Actually this will happen with probability $0.67$ which is indeed big, but not so big. Nevertheless, if $S\ll T$ (i.e. $S$ very small compare to $T$) then indeed $$\mathbb P\{\forall t\in [0,S],-\sigma \sqrt T\leq W_t\leq \sigma \sqrt T \},$$ will be very close to $1$.