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Math Help - Proofs make me cry. Please help

  1. #1
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    Proofs make me cry. Please help

    My questions come from a pdf file and I have no idea how to put the symbols on here, so I took a screenshot and attached my questions as a picture. Can somebody please, please help me out with numbers 4 and 5? I've been trying but I never know what to do with proofs.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by babbagandu View Post
    My questions come from a pdf file and I have no idea how to put the symbols on here, so I took a screenshot and attached my questions as a picture. Can somebody please, please help me out with numbers 4 and 5? I've been trying but I never know what to do with proofs.
    Help make you cry?
    if you insist
    This is not a real proof, it's a calculus calculation
    And you can find this online too.
    By definition

    E(e^{Xt})={1\over \Gamma(\alpha)\beta^{\alpha}}\int_0^{\infty}e^{xt}  x^{\alpha-1}e^{-x/\beta}dx

    NOW combine the exponents of e in the integrand and figure out your NEW beta and you are done.
    Last edited by matheagle; December 3rd 2009 at 10:55 PM.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    Help make you cry?
    if you insist
    This is not a real proof, it's a calculus calculation
    And you can find this online too.
    Oh ok I think I see now. Do you think you could help me out on # 5? I really appreciate your help

    Oh and I am still a little stuck on 4. What do you mean by calculate the new beta?
    Last edited by mr fantastic; December 3rd 2009 at 01:03 PM. Reason: Merged posts
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  4. #4
    MHF Contributor matheagle's Avatar
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    since densities integtrate to one....


    E(e^{Xt})={1\over \Gamma(\alpha)\beta^{\alpha}}\int_0^{\infty}e^{xt}  x^{\alpha-1}e^{-x/\beta}dx


    ={1\over \Gamma(\alpha)\beta^{\alpha}}\int_0^{\infty}x^{\al  pha-1}e^{-x(-t+1/\beta)}dx

    this is infinite unless -t+1/\beta >0

    NOW let 1/\beta_*=-t+1/\beta

    ={\beta_*^{\alpha}\over \Gamma(\alpha)\beta_*^{\alpha}\beta^{\alpha}}\int_  0^{\infty}x^{\alpha-1}e^{-x(-t+1/\beta)}dx

    = {\beta_*^{\alpha}\over \beta^{\alpha}}<br />
\int_0^{\infty}{1\over \Gamma(\alpha)\beta_*^{\alpha}}<br />
x^{\alpha-1}e^{-x(-t+1/\beta)}dx

     ={\beta_*^{\alpha}\over \beta^{\alpha}}=\Biggl({\beta_*\over \beta}\Biggr)^{\alpha}

    using 1/\beta_*=-t+1/\beta

    we have \beta/\beta_*=-t\beta +1

    or \beta_*/\beta={1\over 1-t\beta}

    raise that to the \alpha and you're done
    Last edited by matheagle; December 3rd 2009 at 10:59 PM.
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