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Moment Generating Function

  1. Aug 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Let X be uniformly distributed over the unit interval (0,1). Determine the moment generating function of X, and using this, determine all moments around the origin.

    2. Relevant equations



    3. The attempt at a solution

    I know that the MGT is M(x) = E[ext]

    I'm just not sure how to start this problem. Could someone give me a hint, or point me in the right direction?
     
  2. jcsd
  3. Aug 9, 2009 #2

    Dick

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    That makes it the integral from 0 to 1 of exp(x*t)*1*dx, doesn't it?
     
  4. Aug 9, 2009 #3
    What does the "1" exp(x*t)*1*dx supposed to represent?
     
  5. Aug 9, 2009 #4

    Dick

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    If the distribution is uniform over [0,1], then the pdf is f(x)=1, isn't it? The moment generating function is integral f(x)*exp(x*t)*dx. The 1 is the f(x).
     
  6. Aug 9, 2009 #5
    Can't believe I didn't realise that. Thanks again.
     
  7. Aug 10, 2009 #6
    So I did the integral and I got [tex]\frac{1}{t}e^t - \frac{1}{t}[/tex]

    so the nth derivative of e^t is [tex]\frac{e^t}{t!}[/tex] (from taylor series)

    but not sure what to do about the 1/t.
     
  8. Aug 10, 2009 #7

    Dick

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    The series expansion of e^t is 1+t+t^2/2!+t^3/3!+... Which is quite unlike what you said. The 1/t part cancels.
     
    Last edited: Aug 10, 2009
  9. Aug 10, 2009 #8
    so all moments around the origin is just e^t (evaluated at t = 0)?
     
  10. Aug 11, 2009 #9

    Dick

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    No! Not at all! I was suggesting you expand the infinite series for e^t and use some algebra to write it in a way that's easier to handle for derivatives. You can work directly with e^t/t-1/t as well, but you can't just put t=0 after you take the derivative since that would be undefined - you would have to take the limit as t->0.
     
  11. Aug 11, 2009 #10
    I see what you mean. I get tn - 1/n!
     
  12. Aug 11, 2009 #11

    Dick

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    Good! You mean the sum of that for n=1 to infinity, right? I.e. 1+t/2!+t^2/3!+..., right? Now it's easy to find the nth derivative of that, also right?
     
  13. Aug 11, 2009 #12
    the nth derivative would be [(1-n)^n]*(tn - 1/n!)
     
  14. Aug 11, 2009 #13

    Dick

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    Why on earth would it be that? Look, take 1+t/2!+t^2/3!+t^3/4!+... What's the first derivative evaluated at t=0? Now do the second. Try the third. Do you see a pattern? Those are the moments. They are also pretty easy to compute from the pdf without even using a generating function. You should probably do the zero moment as well.
     
  15. Aug 11, 2009 #14
    probably wrong, but it look's like tn/(n+1)! but if I substitute t = 0, then I get 0.
     
  16. Aug 11, 2009 #15

    Dick

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    Yes, it's wrong. Can you help me out here? m(t)=1+t/2!+t^2/3!+t^3/4!+... That is your moment generating function isn't it? It's a infinite SUM of terms like you have suggested before. Not just a naked t^n/(n+1)!. It's a summation. What's the first derivative at t=0? Please, help me. Exactly ONE of the terms in that series has a nonzero first derivative at t=0. NONE of the others do.
     
    Last edited: Aug 11, 2009
  17. Aug 13, 2009 #16
    So I was looking through my notes and found this:

    [tex]M(0) + \sum M^{(n)}(0)\frac{t^{n-1}}{n!} =1 + \sum^{\infty}_{n=2} E[X^n] \frac{t^{n-1}}{n!}[/tex]

    where E[Xn] = 1/(n+1) since it's uniformly distributed on (0,1)
     
  18. Aug 13, 2009 #17

    Dick

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    Yes. E[X^n]=1/(n+1). Because i) the integral from 0 to 1 of x^n is 1/(n+1). And ii) you can also read that off from the generating faction. If you take the nth derivative and set t=0 the only term that contributes is t^n/(n+1)!. All of the others vanish. And differentiating that term n times gives 1/(n+1).
     
  19. Aug 13, 2009 #18
    So it's supposed to be:

    [tex]M(0) + \sum M^{(n)}(0)\frac{t^{n-1}}{n!} =1 + \sum^{\infty}_{n=2} \frac{1}{n+1} \frac{t^{n-1}}{n!}[/tex]

    Kinda confused about t^n/(n+1)!.
     
  20. Aug 13, 2009 #19

    Dick

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    Forget the n. Just deal with say the first 3 or 4 moments. Try and prove that the first few moments of your distribution are the same as what you would get from the generating function. Or even the first 1 or 2. Any of them. I'm on the verge of giving up here. Just do numbers, forget the symbols.
     
    Last edited: Aug 13, 2009
  21. Aug 13, 2009 #20
    *smacks head on keyboard*

    can't believe it took me so long to get it. Thanks for being so patient, and sorry for all the trouble.
     
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