# Hitting time process of Brownian motion [closed]

I am stuck with this problem:

Let $(B_t)$ be a standard Brownian motion in $\mathbb{R}$. For $t \geq 0$, let $$H_t = \inf \{ s \geq 0 : B_s = t \}, \quad S_t = \inf \{ s \geq 0 : B_s > t \}.$$

I want to prove three things:

$1.$ Given $t \geq 0$, $H_t$ and $S_t$ are a.s. equal. (I tried to use the continuity property of Brownian paths, but still can't show this.)

$2.$ Give an example of a sample path of $\{B_t\}$ such that $\{S_t\}$ and $\{H_t\}$ are not equal on a set of positive measure.

$3.$ Prove that ${S_t}$ is almost-surely nowhere continuous.

Any ideas on how to approach this hard problem?

## closed as off-topic by Did, user147263, Adriano, JimmyK4542, KirillDec 22 '14 at 6:50

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• There seems to be a huge gap between the series of questions you asked recently and your background (leading to essentially unanswerable questions). Please explain. – Did Dec 22 '14 at 6:09

1)As, Brownian paths are continuous its clear that $H_t \leq S_t$ a.s. Also, we know that $B_{\tau + s} - B_{\tau}$ is Brownian Motion for any finite stopping time $\tau$. Now, take $\tau = H_t$ which means $B_{H_t+s}-t$ is Brownian motion. Brownian motion oscillates infinitely at $0$ and hence we can find sequence of times $s_n$ such that $s_n \to 0$ and $B_{H_t+s_n}-t$ are positive and converge to $0$ which means $H_t \geq S_t$ a.s. and hence $H_t = S_t$ a.s.