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Math Help - normal distribution (mgf)

  1. #1
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    normal distribution (mgf)

    Hi

    Can someone please explain the process in creating the moment generating function for the normal distribution? i can do it for the other special distributions but not the normal.

    thanks.
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  2. #2
    Member Last_Singularity's Avatar
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    Quote Originally Posted by gvidfhi View Post
    Hi

    Can someone please explain the process in creating the moment generating function for the normal distribution? i can do it for the other special distributions but not the normal.

    thanks.
    The easiest method to derive the moment-generating function of a general normal distribution N(\mu, \sigma^2) is to find the moment for a standard normal N(0,1) and then use the formula for the linear transformation of a moment. Given (\Omega, F, P), we have a probability space.

    Lemma: Let X: \Omega \longrightarrow R be an absolutely continuous random variable whose moment-generating function is M_x (s). Then if Y = aX +b, then M_y (s) = e^{sb} M_x (sa)

    Proof: Let a second random variable Y = aX + b. Then the moment generating function for Y is M_y (s) = E(e^{sY})
    = E(e^{s(aX+b)})
    = E(e^{saX+sb})
    = E(e^{saX} e^{sb})
    = e^{sb} E(e^{saX})
    = e^{sb} M_x (sa)

    Therefore, if Y = aX +b, then M_y (s) = e^{sb} M_x (sa) - this completes the lemma.

    Consider X, which is standard normally distributed with mean 0 and variance 1, so that f_x (x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}

    The moment generating function for X is calculated by M_x (s) = E(e^{sx})
    = \int_{-\infty}^{\infty} e^{sx} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}
    = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{sx -x^2/2}
    = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} ((x-s)^2 - s^2)}
    = e^{s^2/2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} (x-s)^2}
    = e^{s^2/2}

    Finally, consider Y, which is normally distributed with mean \mu and variance \sigma, so that Y = aX +b = \sigma X + \mu

    Recalling that M_y (s) = e^{sb} M_x (sa) from our lemma, we have M_y (s) = e^{s \mu} M_x (s \sigma) = e^{s \mu} e^{s^2 \sigma^2/2} = e^{s \mu + s^2 \sigma^2/2} \qquad QED
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