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Math Help - Martingale & Stopping time

  1. #1
    Moo
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    Martingale & Stopping time

    Hi !

    Okay, this one has been bugging (and p***ing) me...

    We have (Y_i)_i iid rv's where P(Y_1=1)=p and P(Y_1=-1)=1-p

    Let a\in\mathbb{N}^* and S_n=a+\sum_{i=1}^n Y_i

    So for p=1/2, S_n is a martingale (with the natural filtration)

    Let S_{n\wedge T}=S^T be the stopped martingale, where T=\inf\{n>0 ~:~ S_n=0 \text{ or } S_n=a+b\} (b is another positive integer)

    Then I have to prove, using the convergence of S^T, that P(T<\infty)=1.

    And I don't know how to do it (I've been scrabbling on it, without finding a formal proof...).

    Thanks for any hint...
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  2. #2
    MHF Contributor

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    I think you'll be very disappointed when you get it...

    Hint: almost-surely, S^T is an integer-valued sequence that has a finite limit.
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  3. #3
    Moo
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    I'm disappointed (of myself) because I still don't understand

    1. How do we show that it converges ? The only theorem that we have is this one (I know there are others, though, but I learnt them in another course) :
    Xn is a supermartingale such that sup E|Xn| is finite. Then there exists a rv X such that E|X| is finite and Xn converges to X a.s.
    But here, the condition of S^T to be bounded is unknown because E|S^T|<T or that E|S^T|<n, ce qui ne fait pas avancer le schmilblick

    2. Then I understand clearly why, if S^T converges, then it's constant from a certain rank. But the only way I can see to show this is : if it converges to m\in(0,a+b), it's not possible because the following step is different from m, a.s.. So it must converge to 0 or a+b. Is that correct to write it this way ?


    Thanks, as always
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  4. #4
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    Quote Originally Posted by Moo View Post
    1. How do we show that it converges ? The only theorem that we have is this one (I know there are others, though, but I learnt them in another course) :
    Xn is a supermartingale such that sup E|Xn| is finite. Then there exists a rv X such that E|X| is finite and Xn converges to X a.s.
    But here, the condition of S^T to be bounded is unknown because E|S^T|<T or that E|S^T|<n, ce qui ne fait pas avancer le schmilblick
    Sorry I didn't understand your question also concerned this part. Notice that the martingale S^T is, by construction, bounded: it is stopped when reaching either 0 or a+b, hence it can't leave [0,a+b]...

    2. Then I understand clearly why, if S^T converges, then it's constant from a certain rank. But the only way I can see to show this is : if it converges to m\in(0,a+b), it's not possible because the following step is different from m, a.s.. So it must converge to 0 or a+b. Is that correct to write it this way ?
    This way, you don't really prove that the limit is an integer (I know, this is so obvious...). I would first recall the general result that a convergent integer-valued sequence is stationary (and give a short proof: the sequence is Cauchy, hence we have |u_n-u_m|<1 for m,n\geq n_0, hence u_n=u_m). And then conclude: almost-surely, S^T is stationary; however, S_{T\wedge n}=S_{T\wedge(n+1)}\pm 1 if n< T; therefore almost-surely there exists n\geq T, which is to say that T is finite.
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  5. #5
    Moo
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    Quote Originally Posted by Laurent View Post
    Sorry I didn't understand your question also concerned this part. Notice that the martingale S^T is, by construction, bounded: it is stopped when reaching either 0 or a+b, hence it can't leave [0,a+b]...
    Nah, I'm sorry, I wasn't clear about that.
    And yeah, I actually already thought of that, I just forgot to write it and hence forgot it ...


    This way, you don't really prove that the limit is an integer (I know, this is so obvious...). I would first recall the general result that a convergent integer-valued sequence is stationary (and give a short proof: the sequence is Cauchy, hence we have |u_n-u_m|<1 for m,n\geq n_0, hence u_n=u_m). And then conclude: almost-surely, S^T is stationary; however, S_{T\wedge n}=S_{T\wedge(n+1)}\pm 1 if n< T; therefore almost-surely there exists n\geq T, which is to say that T is finite.
    Nice one

    Now it's all packed, solved, clear. Thanks !

    Moral : don't try to do maths after a too short night
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  6. #6
    Moo
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    Oh, another question (just checking if it's correct) :

    I've showed that S_n'=(S_n-a)^2-n is a martingale.
    And from this, I have to compute E[T].

    Is it correct... ?

    Since it's a martingale, \left((S_{n\wedge T}-a)^2-n\wedge T\right)_n is too.
    So in particular, E[(S_{n\wedge T}-a)^2]-E[n\wedge T]=E[S_0-a]=0

    So \lim_{n\to\infty}E[(S_{n\wedge T}-a)^2]=\lim_{n\to\infty} E[n\wedge T]

    By applying the dominated convergence theorem to the LHS and because T is finite a.s., we have E[(S_T-a)^2]=\lim_{n\to\infty} E[n\wedge T]

    But for the RHS, I don't really know... We can't apply the dominated theorem because we only know that T is finite, not bounded ? I prefer asking, because I know I have many troubles with the bounded stuff...
    But we can apply the monotone convergence theorem, can't we ?

    That's just what I want to know
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  7. #7
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    Quote Originally Posted by Moo View Post
    We can't apply the dominated theorem because we only know that T is finite, not bounded ?
    But we can apply the monotone convergence theorem, can't we ?
    Twice correct!
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