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Math Help - Calculating the MGF

  1. #1
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    Calculating the MGF



    This a pretty weird question... because:

    E(e^{tX}) = M(t) = \int_0^{\infty} e^{xt} e^{-x} dx = \int_0^{\infty} e^{-x(1-t)}dx = \lim_{k \to \infty} \left[\frac{e^{x(t-1)}}{t-1}\right]_0^k

    But the limit: \lim_{k \to \infty} \left[\frac{e^{k(t-1)}}{t-1}\right] is undefined?

    How am I meant to compute the MGF then?

    Thanks
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  2. #2
    GJA
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    Re: Calculating the MGF

    Hi, usagi_killer.

    I'm not familiar with what MGF stand for, but it looks like you may want to re-check your antiderivative step. I think there needs to be a negative in the exponent and a negative out front of everything; doing this, then evaluating the limit in k (assuming t>1) should give the correct result.

    Does this help?

    Good luck!
    Last edited by GJA; August 16th 2012 at 01:45 PM.
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  3. #3
    MHF Contributor harish21's Avatar
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    Re: Calculating the MGF

    Quote Originally Posted by usagi_killer View Post


    This a pretty weird question... because:

    E(e^{tX}) = M(t) = \int_0^{\infty} e^{xt} e^{-x} dx = \int_0^{\infty} e^{-x(1-t)}dx = \lim_{k \to \infty} \left[\frac{e^{x(t-1)}}{t-1}\right]_0^k

    But the limit: \lim_{k \to \infty} \left[\frac{e^{k(t-1)}}{t-1}\right] is undefined?

    How am I meant to compute the MGF then?

    Thanks
    Hi GJA: MGF means moment generating function

    you can write \left[\frac{e^{x(t-1)}}{t-1}\right]_0^k as \frac{1}{-(1-t)} \bigg[e^{-x(1-t)}\bigg]_0^\infty and proceed
    Last edited by harish21; August 16th 2012 at 06:48 PM.
    Thanks from usagi_killer
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  4. #4
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    Re: Calculating the MGF

    Thanks for that,

    However, the above still diverges for t>1 doesn't it?
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  5. #5
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    Re: Calculating the MGF

    I think you are meant to assume t < 1, or deduce that this is a necessary assumption.
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