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Math Help - marginal moment for random sums - question about proof

  1. #1
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    marginal moment for random sums - question about proof

    Suppose that { X_j} is a sequence of intedependent identically distributed random variables and that N is a random variable taking non-negative integer values.

    If Y=\Sigma_{j=1}^N, then M_Y(t)=M_N(lnM_X(t))

    Proof.

    M_Y(t)=E[M_{Y|N}(t|N)]=E[(M_X(t)^N]=E[e^{NXt}]=E[exp({N*ln(e^{xt})]=
    =E(e^{N*lnM_X(t)})=M_N(lnM_X(t))

    The beginning and the end is from the textbook, the part in the middle with E[e^{NXt}]=E[exp({N*ln(e^{xt})] is mine.

    I am not quite sure about the last transition

    E[exp({N*ln(e^{xt})]=E(e^{N*lnM_X(t)})=M_N(lnM_X(t)).

    I can see how E[e^{N*something}] becomes M_N(something). I am not sure why ln(e^{xt})=lnM_X(t) if there is no expected value E anywhere around (since the definition of MGF is M_X(t)=E(e^{Xt}).
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  2. #2
    Moo
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    Hello !
    Quote Originally Posted by Volga View Post
    The beginning and the end is from the textbook, the part in the middle with E[e^{NXt}]=E[exp({N*ln(e^{xt})] is mine.

    (...)

    I can see how E[e^{N*something}] becomes M_N(something). I am not sure why ln(e^{xt})=lnM_X(t) if there is no expected value E anywhere around (since the definition of MGF is M_X(t)=E(e^{Xt}).
    That's because it's E[e^{NXt}]=E[\exp(N*\ln(e^{Xt}))] (capital X)



    P.S. : HK \o///
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  3. #3
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    It's not so obvious to me.

    Instead, I would write E[exp(N*ln(e^{Xt})]=M_N(ln(e^{Xt}))=M_N(Xt)

    ?? I suspect there is something here about the expected value operator that I don't know...
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  4. #4
    Moo
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    Oh right sorry, I misunderstood your question.

    There is a big mistake for the third '=' sign :
    E\left[M_X(t)^N\right]=E\left[\left(E\left[e^{Xt}\right]\right)^N\right], which is completely different from what you wrote.

    I would like to know what you mean by "the beginning" and "the end". Do you mean that the book only provides M_Y(t)=M_N(\ln M_X(t)) ?
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  5. #5
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    Quote Originally Posted by Moo View Post
    Oh right sorry, I misunderstood your question.

    There is a big mistake for the third '=' sign :
    E\left[M_X(t)^N\right]=E\left[\left(E\left[e^{Xt}\right]\right)^N\right], which is completely different from what you wrote.

    I would like to know what you mean by "the beginning" and "the end". Do you mean that the book only provides M_Y(t)=M_N(\ln M_X(t)) ?
    This is from the book:

    M_Y(t)=E[M_{Y|N}(t|N)]=E[(M_X(t)^N]=E(e^{N*lnM_X(t)})=M_N(lnM_X(t))

    The mistake you pointed out is mine ))) I now understand completely, from the double expected value expression you wrote. Thank you!!
    Last edited by Volga; February 1st 2011 at 12:25 AM.
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