Stopping times: [itex]\{min(\sigma,\tau) \leq t\} \in \mathscr{F}_t\}[/itex]

[operationsres]

Homework Statement

Context: Both $\sigma$ and $\tau$ are stopping times in the filtered probability space $(\Omega,\mathscr{F},\{\mathscr{F}_t\}_{t\in [0,\infty)},P)$.

Question: Why is it the case that $\{min(\sigma,\tau) \leq t\} = \{\sigma \leq t\}\cup \{\tau \leq t\}$?

The Attempt at a Solution

I don't know why.

 

[Ray Vickson]

Homework Statement

Context: Both $\sigma$ and $\tau$ are stopping times in the filtered probability space $(\Omega,\mathscr{F},\{\mathscr{F}_t\}_{t\in [0,\infty)},P)$.

Question: Why is it the case that $\{min(\sigma,\tau) \leq t\} = \{\sigma \leq t\}\cup \{\tau \leq t\}$?

The Attempt at a Solution

I don't know why.

To show equality of two events A and B, one way is to show that
$$A \subset B \text{ and } B \subset A.$$
Think about it in *words*: what does the left-hand event say about σ and τ? What does the event on the right say about σ and τ?

RGV

 

[operationsres]

Thanks Ray.

I understand that the two way implications hold when the LHS and RHS are thought of as events (i.e. LHS and RHS are either both true or both not true).

Can you tel lme if the following is correct? I think that this will alleviate some of my confusion.

$\{\sigma \leq t\} \cup \{\tau \leq t\} = \{x \in \mathbb{R} | 0 \leq x \leq t\}$

I think that this expression holds by the fact that both $\sigma,\tau \geq 0$ are random variables. So the event of these random variables being $\leq t$ is simply all the real numbers from $0$ to $t$ (inclusive of 0 and t).