Continuity of Passage Time

Question

Let $\left\{ W_{t},\mathscr{F}_{t}:0\le t<\infty\right\}$ be a standard, one-dimensional Brownian motion on $\left(\Omega,\mathscr{F},P\right)$, and define $$T_{b}=\inf\left\{ t\geq0:W_{t}=b\right\} ;\quad b\geq0$$ $$S_{b}=\inf\left\{ t\geq0:W_{t}>b\right\} ;\quad b\geq0$$ Then how can we show that for fixed $\omega\in\Omega$, $T_{b}\left(\omega\right)$ is a left-continuous function of $b$ and $S_{b}\left(\omega\right)$ is a right-continuous function of $b$? Can anyone help me? Thanks a lot in advance!

Answer 1

For almost all $\omega$, $W_t$ is continuous in $t$.

First, this implies $T_b=\min\{t: W_t=b\}$. So if $b'<b$, by intermediate value theorem $T_{b'}< T_b$. On the other hand, for any $\epsilon>0$, there is $\delta>0$ so that $\sup\{W_t: t\in [0, T_n-\epsilon]\}\leq b-\delta$, thus if $b'>b-\delta$ we see $T_{b'}\geq T_b-\epsilon$, so $\lim_{b'\to b-}T_{b'}=T_b$.

Next, it is easy to see if $b'>b$ then $S_{b'}\geq S_b$. On the other hand, for any $\epsilon>0$ there is $\delta$, so that there is $t\in [S_b, S_b+\epsilon]$ so that $W_t>b+\delta$. So if $b'\in [b, b+\delta]$, we see $S_{b'}\leq S_b+\delta$. This proves $\lim_{b'\to b+}S_{b'}=S_b$.