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Math Help - de Moivre's martingale

  1. #1
    Moo
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    de Moivre's martingale

    Hi !

    Looks like it's the same as last year : I can't wait for the next class to solve a problem ^^'
    We've studied martingales for only 3 hours already, so I'm still very new to it.

    So we have a sequence (Y_n)_{n\geq 1} of iid rv's such that P(Y_n=-1)=q and P(Y_n=1)=p
    where p+q=1 \ , \ 0<p<q<1
    Let X_0=0 \ , \ Z_0=1... X_n=Y_1+\dots+Y_n and Z_n=\left(\tfrac qp\right)^{X_n}

    Let's consider the filtration \mathcal{F}_n=\sigma(Z_0,\dots,Z_n)

    I have to show that Z_n is a positive martingale.

    3 points to show :
    1/ Z_n is \mathcal{F}_n-measurable
    2/ Z_n is integrable
    3/ E(Z_{n+1}\mid \mathcal{F}_n)=Z_n

    Point 1/ is okay.
    Point 2/ is okay.
    Point 3/ is totally unknown Which method to use ?


    Thanks for any help/hints
    Last edited by Moo; October 19th 2009 at 09:03 AM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Fine I'll do it. Conditioning on the Z's you can pull the first n Y's out of the expectation...

    \biggl({q\over p}\biggr)^{Y_1+\cdots +Y_n} E\biggl({q\over p}\biggr)^{Y_{n+1}}

    Now use your distribution of the Y's

    E\biggl({q\over p}\biggr)^{Y_{n+1}} =\biggl({q\over p}\biggr)^{-1}q +\biggl({q\over p}\biggr)^1p=p+q=1

    So your conditional expectation is

    \biggl({q\over p}\biggr)^{Y_1+\cdots +Y_n}=\biggl({q\over p}\biggr)^{X_n}=Z_n

    Last edited by matheagle; October 20th 2009 at 10:02 PM. Reason: typo with X's and Z's
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  3. #3
    Moo
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    Darn ! That was really...

    Thanks !
    Last edited by mr fantastic; October 20th 2009 at 02:08 PM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    easy?
    Everything is easy when you know what you're doing.
    I used to say certain problems were easy (and also some were hard)
    in class and that would annoy the students who didn't realize the difference.

    But I don't know why they have the condition p<q here.
    Maybe that's the next step.
    That we expect more failures than successes.
    This then goes to infinity.

    and I saw the unedited post, lol.
    I've found out that the cow sees everything.

    I have exams to grade next week, if anyone is interested.
    Last edited by matheagle; October 20th 2009 at 04:43 PM. Reason: But did moo see this one ....?
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  5. #5
    Moo
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    Quote Originally Posted by matheagle View Post
    easy?
    Everything is easy when you know what you're doing.
    I used to say certain problems were easy (and also some were hard)
    in class and that would annoy the students who didn't realize the difference.
    Hmmm no, I would've said obvious or direct. Once you've done enough of conditional expectation of course.

    But I don't know why they have the condition p<q here.
    Maybe that's the next step.
    Yep, we had to prove then that X_n goes to -infinity. And if you're interested, the following questions are :

    2. Let T_k=\inf\{n\geq 0 : X_n\geq k\}
    Consider the martingale Z_{T_k\wedge n} (it's true because Tk is a bounded stopping time and Zn is a martingale)
    Then show that P(T_k<\infty)=\left(\tfrac pq\right)^k
    (I think this is where the condition over p and q is the most important actually)

    3. Finally show that E(\sup_{n\geq 0} X_n)=\frac{p}{q-p}

    and I saw the unedited post, lol.
    I've found out that the cow sees everything.
    Huh ? which undedited post ?
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  6. #6
    MHF Contributor matheagle's Avatar
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    finals are over!!!
    Last edited by matheagle; December 15th 2009 at 08:54 PM.
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