1. ## martingale stopping time

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

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

2. What do you mean by this?

3. Hello,

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

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

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

QED

4. Originally Posted by Moo
Hello,

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

Now we have to check that $\displaystyle \{S+T\leq n\}\in \mathcal F_n$, which is straightforward since $\displaystyle \{S+T\leq n\}=\bigcup_{t=0}^n \{S\leq n-t\}$
Slight error here. It should be $\displaystyle \{S \leq n-t\}\cap\{T\leq t\}$ in the union. Hence why you need both T and S to be stopping times.

5. Focus,

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

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

6. Originally Posted by Moo
Focus,

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