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Thread: martingale stopping time

  1. #1
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    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?
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  2. #2
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    What do you mean by this?
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  3. #3
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    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
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    Quote Originally Posted by Moo View Post
    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.
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  5. #5
    Moo
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    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 $\displaystyle \mathcal F_n$
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    Quote Originally Posted by Moo View Post
    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 $\displaystyle \{S\leq n-t\}\cap\{T \leq t\}\subset \{S+T \leq n\}$.
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