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Thread: martingale proof

  1. #1
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    martingale proof

    I need to do the proof for the second condition but I'm not getting the right answer.

    $\displaystyle \{S_n , n\geq 1\}$} is a martingale if

    1) E[|$\displaystyle S_n$|]< $\displaystyle \infty \forall n$
    2) E[$\displaystyle S_{n+1}|S_1,...,S_n$] = $\displaystyle S_n \forall n$

    $\displaystyle S_n$=$\displaystyle X_1X_2...X_n$

    Suppose E[|$\displaystyle X_n$|] < $\displaystyle \infty \forall n$ and E[$\displaystyle X_n$]=1

    Prove that $\displaystyle \{S_n , n\geq 1\}$} is a martingale



    I can prove the first condition but for the second one I keep getting E[$\displaystyle S_{n+1}|S_1,...,S_n$] = $\displaystyle S_n+1$ which is wrong. Can anyone shine a light on this problem? Thanks

    (ps sorry for any bad use of LaTex, this is the first time I have used it!)
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by willowtree View Post
    I need to do the proof for the second condition but I'm not getting the right answer.

    $\displaystyle \{S_n , n\geq 1\}$} is a martingale if

    1) E[|$\displaystyle S_n$|]< $\displaystyle \infty \forall n$
    2) E[$\displaystyle S_{n+1}|S_1,...,S_n$] = $\displaystyle S_n \forall n$

    $\displaystyle S_n$=$\displaystyle X_1X_2...X_n$

    Suppose E[|$\displaystyle X_n$|] < $\displaystyle \infty \forall n$ and E[$\displaystyle X_n$]=1

    Prove that $\displaystyle \{S_n , n\geq 1\}$} is a martingale



    I can prove the first condition but for the second one I keep getting E[$\displaystyle S_{n+1}|S_1,...,S_n$] = $\displaystyle S_n+1$ which is wrong. Can anyone shine a light on this problem? Thanks

    (ps sorry for any bad use of LaTex, this is the first time I have used it!)
    If this is a product and not a sum, then

    E[$\displaystyle S_{n+1}|S_1,...,S_n] = E( X_1X_2\cdots X_{n+1}|S_1,...,S_n)=X_1X_2\cdots X_nE(X_{n+1}|S_n)$

    and I'm guessing you need $\displaystyle E(X_1)=1$

    YUP, I see it there.
    I found your $\displaystyle E(X_1)=1$

    $\displaystyle =X_1X_2\cdots X_nE(X_{n+1}) =S_n (1)=S_n $
    Last edited by matheagle; May 5th 2010 at 05:05 PM.
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  3. #3
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    ahh i seee, thank you, i was doing something a bit backwards with the stuff inside the brackets!
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