Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am new here and I have a question.

Defenition: Let $ \tau$ be a stopping time, then $\digamma_{\tau}=\{F\subset \Omega: \forall n \in N \cup \{\infty\} , F\cap(\tau\leq n)\in \digamma_{n}$} is a sigma-algebra.

Then my question is: can somebody prove that

$ \digamma_{\tau_{1} \cap \tau_{2}}=\digamma_{\tau_{1}} \cap \digamma_{\tau_{2}}$,where $\tau_{1}$ and $\tau_{2}$ are both stopping times and $\tau_{1} \wedge \tau_{2}=min(\tau_{1},\tau_{2})$.

My attempt to prove this:

Since $\forall F \in \digamma_{n},$ $F\cap(\tau\leq n)\in \digamma_{n}$ iff $F\cap(\tau\leq n)^{c}\in \digamma_{n}$, we can write

$\digamma_{\tau}=\{F\subset \Omega: \forall n \in N \cup \{\infty\} , F\cap(\tau > n)\in \digamma_{n}\} $.

Then we can write $ \digamma_{\tau_{1} \cap \tau_{2}}=\{F\subset \Omega: \forall n \in N \cup \{\infty\} , F\cap(\tau_{1}\leq n)\cap(\tau_{2} \leq n)\in \digamma_{n}\} = \digamma_{\tau_{1} \cap \tau_{2}}$

Since $(min(\tau_{1},\tau_{2})\leq n)=(\tau_{1}\leq n)\cap(\tau_{2} \leq n)$.

My question is now, is this prove legit?

share|improve this question
1  
The system works a lot better if you only ask one question per question. And if you want to get useful answers then it's helpful if you can indicate what you've tried and why it didn't work. –  Peter Taylor Jun 25 '13 at 21:25
    
agreed ^ i am trying to read this whole thing but it is so long. so you have a latex error with the 'and' in question 1, i cant fix it cos it is too minor a correction. but it annoys me when i read it –  Lost1 Jun 25 '13 at 22:35
    
@Peter Taylor OK, I'll edit this question, and post multiple questions then –  Niels Jun 26 '13 at 12:52
add comment

2 Answers 2

I actually wanted to delete this post after I have written it, because I have seen no evidence that you tried anything yourself. However, I will give you the benefit of doubt. These are some food for thought, but I cannot guarantee they will work. If I was candid, I don't like doing other people's homework, unless I can learn a great deal from it. Also, such questions are not welcome on this site.

For question 1: Here is an idea. For your defintion of $\mathcal{F}_\tau$. I think it is equivalent to $F\in\Omega... F\bigcap \{\tau>n\}\in \mathcal{F}_n$. Now try to write $\min \{\tau_1,\tau_2\}>n$ as the intersection of the events $\tau_1>n$ and $\tau_2>n$. I hope this would work.

For question 2:

you write $M_\tau = \sum\limits_{i=0}^\infty 1_{\{\tau = i\}} M_i$. Can you show that this is measurable with respct to $\mathcal{F}_\tau$. Note you cannot do this trict if time is continuous

Question 3 is the proof of Optional Sampling Theorem in discrete setting. Look this up somewhere. what you need to do is precisely follow the hint.

Question 4 states a stopped martingale is a martingale.

The hints in question 3 and 4 are reasonable hints. maybe you should try them and post your own answers then people would be willing to help.

share|improve this answer
add comment

Unfortunately, your proof doesn't work out. You proved $\mathcal{F}_{\tau_1 \wedge \tau_2} = \mathcal{F}_{\tau_1 \wedge \tau_2}$ - but that's not really surprising. Note that equation

$$\{\tau_1 \wedge \tau_2 \leq n\} = \{\tau_1 \leq n\} \cap \{\tau_2 \leq n\}$$

does not hold.

So here is a proof: Let $F \in \mathcal{F}_{\tau_1} \cap \mathcal{F}_{\tau_2}$. Since

$$\{\tau_1 \wedge \tau_2 \leq n\} = \{\tau_1 \leq n\} \cup \{\tau_2 \leq n\}$$

we have

$$F \cap \{\tau_1 \wedge \tau_2 \leq n\} = \underbrace{(F \cap \{\tau_1 \leq n\})}_{\in \mathcal{F}_n} \cup \underbrace{(F \cap \{\tau_2 \leq n\})}_{\in \mathcal{F}_n} \in \mathcal{F}_n$$

which proves $F \in \mathcal{F}_{\tau_1 \wedge \tau_2}$. On the other hand, for $F \in \mathcal{F}_{\tau_1 \wedge \tau_2}$ we have

$$F \cap \{\tau_1 \leq n\} \stackrel{\tau_1 \geq \tau_1 \wedge \tau_2}{=} F \cap \{\tau_1 \geq \tau_1 \wedge \tau_2\} \cap \{\tau_1 \leq n\} = F \cap \{\tau_1 \wedge \tau_2 \leq n\} \cap \{\tau_1 \leq n\} \in \mathcal{F}_n$$

Consequently, we proved

$$\mathcal{F}_{\tau_1} \cap \mathcal{F}_{\tau_2} = \mathcal{F}_{\tau_1 \wedge \tau_2}$$

share|improve this answer
    
:As my knowledge, $\tau_1 \wedge \tau_2= \min(\tau_1, \tau_2).$ –  Zbigniew Aug 4 '13 at 10:27
    
@Zbigniew Exactly. What's your point? Note: I wrote that the first equality does not hold! (And yes, sorry, there was a typo in the last equation-line. If that's what you meant: Thanks for the hint.) –  saz Aug 5 '13 at 15:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.