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Thread: Martingale

  1. #1
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    Martingale

    He guys,

    I want to prove that for a sequence $\displaystyle X_{1},X_{2},... $ of independent random variables which have a standard normal distribution, that $\displaystyle Y_{n}=Exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n\right) $ is a martingale.
    I have that $\displaystyle 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:

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

    To go from (2) to (3), notice that conditionally the stuff we factored out is a constant, and is exactly equal to $\displaystyle Y_n$. To go from (3) to (4), we notice that $\displaystyle X_{n + 1}$ 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 $\displaystyle E(\exp\left(X_{n+1}-1/2)) = 1$ 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 $\displaystyle X\sim N(0,1)$ we have $\displaystyle M(t)=E(e^{Xt})=e^{t/2}$

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