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Math Help - martingale stopping time

  1. #1
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    martingale stopping time

    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?
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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 \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
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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 \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.
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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 \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
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  6. #6
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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 \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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