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Math Help - moment generating functions for independent var.

  1. #1
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    moment generating functions for independent var.

    let X~N(mu,sigma^2),Y~Gamma(alpha,Beta) . X and Y are independent , the moment generating function is e^(ut + sigma^2*t^2)/2 for X and (1-Beta)^-alpha for Y. Find mgf of X+Y....I know if you have two indp. var. you multiply them to get the answer, but what form do you leave it in. My answer is e^(ut + sigma^2*t^2)/2 ((1-Beta)^-alpha . Can this answer be broken down anymore?

    2)Also how do you compute mgf for 3X? Is it just 3(e^(ut + sigma^2*t^2)/2) or do you find the differiate 3 times?

    3) last question computing mgf for these types of problems 2+ Y or 2 + 2X + 3Y. Is it just 2 + (1-Beta)^ -alpha and 2 + (3X)*(3Y)?
    I just need to make sure I understand how to compute the mgf.
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  2. #2
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    Hello,
    Quote Originally Posted by pastadee View Post
    let X~N(mu,sigma^2),Y~Gamma(alpha,Beta) . X and Y are independent , the moment generating function is e^(ut + sigma^2*t^2)/2 for X and (1-Beta)^-alpha for Y. Find mgf of X+Y....I know if you have two indp. var. you multiply them to get the answer, but what form do you leave it in. My answer is e^(ut + sigma^2*t^2)/2 ((1-Beta)^-alpha . Can this answer be broken down anymore?
    It's rather
    the moment generating function is e^(ut + sigma^2*t^2/2) for X
    2)Also how do you compute mgf for 3X? Is it just 3(e^(ut + sigma^2*t^2)/2) or do you find the differiate 3 times?
    The definition of the mgf is E(e^{Xt})
    So the mgf of 3X is E(e^{3Xt})=E(e^{X(3t)}), which is the mgf of X taken at the point 3t.

    3) last question computing mgf for these types of problems 2+ Y or 2 + 2X + 3Y. Is it just 2 + (1-Beta)^ -alpha and 2 + (3X)*(3Y)?
    I just need to make sure I understand how to compute the mgf.
    Same method as question 2) : write it in the form of the exponential.
    Try it
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