martingale stopping time

**noob mathematician** · November 26th 2010, 09:50 PM

With S and T the stopping times for a sequence of $\sigma$ -algebra $(F_n)_{n\geq 0}$ , with $F_m \subset F_n$ for m<n.

How can I show that S+T is a stopping time?

**Quantum** · November 27th 2010, 06:25 AM

What do you mean by this?

**Moo** · November 27th 2010, 08:33 AM

Hello,

You just have to use the definition of a stopping time...
Assuming S and T take their values in $\mathbb N$ , S+T does too.

Now we have to check that $\{S+T\leq n\}\in \mathcal F_n$ , which is straightforward since $\{S+T\leq n\}=\bigcup_{t=0}^n \{S\leq n-t\}$

Since S is a stopping time, $\{S\leq n-t\}\in\mathcal F_{n-t}\subset \mathcal F_n$ . And since $\mathcal F_n$ is a sigma-algebra, a finite union of elements of $\mathcal F_n$ will still be in $\mathcal F_n$ .

QED

**Focus** · November 27th 2010, 01:01 PM

Originally Posted by Moo

Hello,

You just have to use the definition of a stopping time...
Assuming S and T take their values in $\mathbb N$ , S+T does too.

Now we have to check that $\{S+T\leq n\}\in \mathcal F_n$ , which is straightforward since $\{S+T\leq n\}=\bigcup_{t=0}^n \{S\leq n-t\}$

Slight error here. It should be $\{S \leq n-t\}\cap\{T\leq t\}$ in the union. Hence why you need both T and S to be stopping times.

**Moo** · November 28th 2010, 04:19 AM

Focus,

Yes it is indeed wrong, but I don't agree with yours either. It should be $\bigcup_{t=0}^n \{S\leq n-t\}\cap \{T=t\}$

By noticing that $\{T=t\}=\{T\leq t\}\cap \{T\leq t-1\}^c$ , it's just a bit longer but not more complicated to show that it belongs to $\mathcal F_n$

**Focus** · November 28th 2010, 07:32 AM

Originally Posted by Moo

Focus,

Yes it is indeed wrong, but I don't agree with yours either. It should be $\bigcup_{t=0}^n \{S\leq n-t\}\cap \{T=t\}$

Those two are identical. Infact it is easy to see that yours is a subset of what I have written. The other inclusion is trivial $\{S\leq n-t\}\cap\{T \leq t\}\subset \{S+T \leq n\}$ .

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