Stopping times, Filtration, Martingales,

Question

I am new here and I have a question.

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

Then my question is: can somebody prove that

$ \mathcal F_{\tau_{1} \wedge \tau_{2}}=\mathcal F_{\tau_{1}} \cap \mathcal F_{\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 $\forall F \in \mathcal F_{n},$ $F\cap(\tau\leq n)\in \mathcal F_{n}$ iff $F\cap(\tau\leq n)^{c}\in \mathcal F_{n}$, we can write

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

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

Answer 1

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.

Answer 2

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

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