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

  • Thread starter operationsres
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In summary, the question is asking about the equality of two events involving stopping times \sigma and \tau in a filtered probability space. The solution involves showing that the left-hand event implies the right-hand event and vice versa. The events can also be thought of as the set of real numbers from 0 to t, inclusive of 0 and t.
  • #1
operationsres
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Homework Statement



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

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


The Attempt at a Solution



I don't know why.
 
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  • #2
operationsres said:

Homework Statement



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

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


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
[tex] A \subset B \text{ and } B \subset A.[/tex]
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
 
  • #3
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.

[itex]\{\sigma \leq t\} \cup \{\tau \leq t\} = \{x \in \mathbb{R} | 0 \leq x \leq t\}[/itex]

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

What is a stopping time?

A stopping time is a random variable that represents the time at which a certain event occurs in a stochastic process. It can take on different values for different outcomes, and must be measurable with respect to the information available at that time.

What is the significance of stopping times in stochastic processes?

Stopping times are important in stochastic processes because they provide a way to analyze and predict when certain events will occur. They also allow for the calculation of probabilities and expected values at specific points in time, which can be useful in decision-making and risk management.

What is the difference between a stopping time and a deterministic time?

A stopping time is a random variable that depends on the outcomes of a stochastic process, while a deterministic time is fixed and independent of the process. Stopping times can vary for different outcomes, while deterministic times are constant.

Can a stopping time be negative?

Yes, a stopping time can be negative. This means that the event it represents has already occurred before time t. For example, if the stopping time represents the time at which a coin lands on heads, a negative stopping time would indicate that the coin landed on heads before time t.

How do you calculate the probability of a stopping time?

The probability of a stopping time can be calculated using the formula P(min(σ,τ) ≤ t) = P(σ ≤ t) + P(τ ≤ t) - P(σ ≤ t, τ ≤ t), where σ and τ are two stopping times. This formula takes into account the different outcomes of the process and the information available at time t.

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