First exit time of Brownian motion

Question

I am new to stopping times and would appreciate any explanation on the following problem:

Let $τ(ω) =$ inf$\{t>0:|W_t(ω)| \geq K\}$, with $K > 0$ constant, be the stopping time. $W_t$ is the Brownian motion.

1. What does the following equation mean?
2. What is the intuition idea behind the following equation? (i.e. why is it true?)

$$\{ω : τ(ω)\leq t\} = \bigcup_{0<s<t}\{ω:|W_s(ω)|\geq K\}$$

This is what I gathered so far:

Since $τ(ω) =$ inf$\{t>0:|W_t(ω)| \geq K\}$, \begin{align} \{ω:τ(ω) \leq t\} &= \tag{1}\left\{ω: \mathrm{inf}\{s>0:|W_s(ω)| \geq K\} \leq t \right\}\\ &= \tag{2}\bigcup_{0<s<t}\{ω:|W_s(ω)|\geq K\} \end{align}

3. How does $(1)$ equal $(2)$?
4. How did the union come about and what happened to the infimum?

EDIT : Despite the 2 answers given, I still have trouble understanding.

Thank you.

Answer 1

Equivalent are the statements:

• $\omega\in\bigcup_{s\in\left(0,t\right)}\left\{ |W_{s}|\geq K\right\} $

• For some $s\in\left(0,t\right)$ we have $|W_{s}\left(\omega\right)|\geq K$

• $\inf\left\{ s>0\mid |W_{s}\left(\omega\right)|\geq K\right\} <t$

• $\tau\left(\omega\right)<t$

Answer 2

The stopping time $\tau(\omega)$ of one particular path $\omega$ is the time this path is further than $K$ away from the origin for the first time.

The line you are interested in says the collection of all paths with stopping time smaller than $t$ is the union over all $s$ smaller than $t$ of all path that at time $s$ are further than $K$ from the origin.

Edit: The left side of the equation is the collection of all paths that are further away than $K$ from the origin at least once before time $t$. So every path in this set needs to be further away than $K$ at some point $s$ with $0 < s < t$. For a fixed $s$, the term in brackets on the right side is the collection of paths that are further away than $K$ at time $s$. The union over all $s$ is exactly the collection of paths that are further away then $K$ at some point between $0$ and $t$.