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MGF and moments

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the MGF and all the moments for [itex]X\sim N(0,1)[/itex]

    2. The attempt at a solution

    For the MGF, I have:

    [tex]M_X(s)=\displaystyle\int_{-\infty}^{\infty}e^{sx}\frac{e^{x^2/2}}{\sqrt{2\pi}}\,dx = \ldots=e^{s^2/2}[/tex]

    Next I found that:
    [tex]M'_X(0)=E[X]=0[/tex]
    [tex]M''_X(0)=E[X^2]=1[/tex]
    [tex]E[X^3]=0[/tex]
    [tex]E[X^4]=3[/tex]
    [tex]\ldots[/tex]
    [tex]E[X^{ODD}]=\{0\}[/tex]
    [tex]E[X^{EVEN}]=\{1,3,15,105,945,\ldots\}[/tex]
    Is it enough to write:
    [tex]E[X^k]=M_X^{(k)}(0)=\frac{d^k}{ds^k}e^{s^2/2}[/tex]

    Am I totally off track here? How would I prove this?
     
  2. jcsd
  3. Feb 9, 2012 #2

    LCKurtz

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    It looks OK to me. How to prove it...it is a general property of Moment Generating Functions and the reason they are called that. If$$
    M(s) = E(e^{sx}) = \int_{-\infty}^{\infty}e^{sx}f(x)\, dx$$and you differentiate with respect to s you get:$$
    M'(s) = \int_{-\infty}^{\infty}xe^{sx}f(x)\, dx$$If you evaluate that at ##s=0## you get$$
    M'(0)=\int_{-\infty}^{\infty}xf(x)\, dx = E(X)$$Each time you differentiate with respect to ##s## you get another ##x## out in front giving you the next moment.
     
  4. Feb 9, 2012 #3

    Ray Vickson

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    Your MGF formula is seriously wrong: you need [itex] e^{-x^2/2}[/itex] in the integrand, not your [itex] e^{x^2/2}.[/itex] However, your result
    [tex]M_X(s) = e^{s^2/2} [/tex] is correct, as are your subsequent results.

    As for how to prove it: just use standard theorems about differentiation under the integral sign. The normal density goes to 0 quickly enough for large |x| that you will not have any problems meeting the hypotheses of the required theorems.

    RGV
     
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