Proving variance with moment generating functions

  • Thread starter TelusPig
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  • #1
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Main Question or Discussion Point

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]
 

Answers and Replies

  • #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?
 
  • #3
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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]
 
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  • #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.
 
  • #5
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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?
 
  • #6
Office_Shredder
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No, you need to plug in t=0!
 
  • #7
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No, you need to plug in t=0!
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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