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Math Help - Martingale

  1. #1
    Junior Member
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    Martingale

    He guys,

    I want to prove that for a sequence  X_{1},X_{2},... of independent random variables which have a standard normal distribution, that  Y_{n}=Exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n\right) is a martingale.
    I have that  E(Y_{n+1}\right)\left|Y_{n}\right)=E(Exp\left(X_{n  +1}-1/2)Exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n\right\right)|Exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n \right)\right).
    What can I do know?

    gr,

    sung
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  2. #2
    Guy
    Guy is offline
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    Given that I know nothing about martingales, take this with a little bit of salt:

    E(Y_{n+1}\right)\left|Y_{n}\right) (1)
    =E(\exp\left(X_{n+1}-1/2)\exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n\right\right)|\exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n  \right)\right). (2)
    = \exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n  \right)\right) E(\exp\left(X_{n+1}-1/2)|\exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n  \right)\right)<br />
(3)
    = Y_{n} E(\exp\left(X_{n+1}-1/2))<br />
(4)
    =Y_n (5)

    To go from (2) to (3), notice that conditionally the stuff we factored out is a constant, and is exactly equal to Y_n. To go from (3) to (4), we notice that X_{n + 1} is independent of the rest of the data, so that it is independent of a function of the rest of the data, so we can get rid of the conditional part of the expectation. To go from (4) to (5), we just grind out that E(\exp\left(X_{n+1}-1/2)) = 1 which is really easy to do by working with the pdf.
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  3. #3
    MHF Contributor matheagle's Avatar
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    NO grinding is necessary

    Since X\sim N(0,1) we have M(t)=E(e^{Xt})=e^{t/2}

    Just let t=1 and you have E(e^{X})=e^{1/2}
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