Results 1 to 4 of 4

Math Help - Deriving MGF of Gaussian distribution

  1. #1
    Newbie Eight^squared's Avatar
    Joined
    Dec 2008
    Posts
    1

    Deriving MGF of Gaussian distribution

    Hi there,
    I'm revising at the moment for a stats exam in about a week's time and I'm having trouble deriving the MGF of the Normal distribution.

    I know that i should get:
    <br />
\exp\left(\mu t +\frac{t^2 \cdot \sigma^2}{2} \right)

    but i just can't seem to get my integration to work.

    Any help would be greatly appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Last_Singularity's Avatar
    Joined
    Dec 2008
    Posts
    157
    Quote Originally Posted by Eight^squared View Post
    Hi there,
    I'm revising at the moment for a stats exam in about a week's time and I'm having trouble deriving the MGF of the Normal distribution.

    I know that i should get:
    <br />
\exp\left(\mu t +\frac{t^2 \cdot \sigma^2}{2} \right)

    but i just can't seem to get my integration to work.

    Any help would be greatly appreciated!
    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

    Sorry, I know that you wanted the variables in terms of t instead of s but s is the convention that I am used to, haha...
    Last edited by Last_Singularity; December 30th 2008 at 05:03 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2010
    Posts
    1
    Quote Originally Posted by Last_Singularity View Post
    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}1
    = e^{s^2/2}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

    Sorry, I know that you wanted the variables in terms of t instead of s but s is the convention that I am used to, haha...
    Last_Singularity: Thx for the explanation, I do hv still one more small problem and thats how do you go from 1 to 2? (see red above)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Last_Singularity's Avatar
    Joined
    Dec 2008
    Posts
    157
    \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} (x-s)^2}
    is simply taking the integral of a normal distribution (with variance 1 and its mean at s) across its domain, or finding its area. And recall that the area under any pdf is just one.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Deriving expected value and variance of beta distribution
    Posted in the Advanced Statistics Forum
    Replies: 11
    Last Post: February 15th 2011, 12:32 AM
  2. Student T-Distribution or Gaussian Distribution
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: September 13th 2010, 01:58 PM
  3. Deriving Beta-distribution from F-distribution
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: February 15th 2010, 03:53 PM
  4. Gaussian Distribution
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 27th 2007, 01:42 PM
  5. Gaussian distribution
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: November 28th 2006, 01:30 PM

Search Tags


/mathhelpforum @mathhelpforum