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Math Help - Discrete Geometric Moment Generating Function

  1. #1
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    Discrete Geometric Moment Generating Function

    Hi everybody,

    I have a question regarding how to proceed with the following problem. I am required to produce a moment generating function for the following problem. I would like to know if the operation I performed is valid for this question. I never understood mgf's that well.

    Question:
    A balanced coin is flipped repeatedly until a head appears. Let X denote the number of flips needed. Define Y = 2X-1.

    Find the moment generating function for Y.

    My reasoning:
    This is a geometric distribution (discrete random variable).

    I applied the definition of the m.g.f.

    m_{y}(t) = \Sigma  e^t(2X-1) \times q^x-1 \times p

    Note: I'm new to latex. Please consider the (2X-1) as raised to the power of e, along with the t.

    Is this the valid first step for this problem, and others like it?

    Thanks for your time!
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  2. #2
    MHF Contributor matheagle's Avatar
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    I do not understand your MGF at all.

    M_Y(t) =E(e^{Yt})  =E(e^{(2X-1)t}) =e^{-t} E(e^{2Xt}) =e^{-t} E(e^{X(2t)}) =e^{-t} M_X(2t)
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  3. #3
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    Quote Originally Posted by matheagle View Post
    I do not understand your MGF at all.

    M_Y(t) =E(e^{Yt}) =E(e^{(2X-1)t}) =e^{-t} E(e^{2Xt}) =e^{-t} E(e^{X(2t)}) =e^{-t} M_X(2t)
    And the mgf for a geometric distribution has been done, among other places, here: http://www.mathhelpforum.com/math-he...tribution.html
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  4. #4
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    Quote Originally Posted by matheagle View Post
    I do not understand your MGF at all.

    M_Y(t) =E(e^{Yt})  =E(e^{(2X-1)t}) =e^{-t} E(e^{2Xt}) =e^{-t} E(e^{X(2t)}) =e^{-t} M_X(2t)
    Haha, boy was I off. Thanks matheagle. I had that in my notes, as I'm guessing that's what my prof did on the blackboard, but I couldn't see how it applied to this problem. This stuff is confusing.

    Thanks
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