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Proving variance with moment generating functions

  1. May 26, 2013 #1
    Moment generating functions:
    How can I show that [itex]Var(X)=\frac{d^2}{dt^2}ln M_X(t)\big |_{t=0}[/itex]

    Recall:
    [itex]M_X(t)=E(e^{tx})=\int_{-\infty}^{\infty}e^{tx}f(x)dx[/itex]

    [itex]E(X^n)=\frac{d^n}{dt^n}M_X(t)\big |_{t=0}[/itex]

    [itex]Var(X)=E(X^2)-[E(X)]^2=E[(X-E(X))^2][/itex]
    ------------
    I tried just applying the equation given but I don't know what to do with the log of this general integral?
    [itex]\frac{d^2}{dt^2}ln M_X(t)=\frac{d^2}{dt^2}ln \left( \int_{-\infty}^{\infty}e^{tx}f(x)dx \right)[/itex]
     
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  3. May 26, 2013 #2

    Office_Shredder

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    If you were asked the more general question of calculating
    [tex] \frac{d^2}{dt^2} \ln\left( f(t) \right) [/tex]
    What's the first thing you would do?
     
  4. May 26, 2013 #3
    It would be for 1st derivative: [itex]\frac{1}{f(t)}*f'(t)[/itex]
    then differentiate again for the 2nd derivative, that would be:

    [itex]\frac{f''(t)f(t)-(f'(t))^2}{(f(t))^2}[/itex]
     
    Last edited: May 26, 2013
  5. May 26, 2013 #4

    Office_Shredder

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    So once you do that you've expressed everything in terms of the generating function and its derivatives without any logarithms involved.
     
  6. May 26, 2013 #5
    Does that mean it's [itex]\frac{M_X''(t)M_X(t)-(M_X'(t))^2}{(M_X(t))^2}[/itex]

    Do I have to then substitute each M(t), M'(t), M''(t) with it's integral definition then? and somehow simplify that big mess o.o?
     
  7. May 26, 2013 #6

    Office_Shredder

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    No, you need to plug in t=0!
     
  8. May 26, 2013 #7
    OH obviously!! LOL omg, I can't believe I didn't see that and thought I had to do a bunch of integrals ._. Thanks!! :D
     
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