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Probability, Moment-Generating Functions

  1. Jul 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose that for a random variable X, E(X^n)=2^n, for n=1,2,3,...through infinity. Calculate the moment generating function (Mx(t)) and the probability mass function (p(i)).


    2. Relevant equations



    3. The attempt at a solution
    It seems as though, letting t=0, M'x(0)=E(X)=2; taking the integral of M'x(0)=2 from 0 to infinity, Mx(t) probably equals 2x (although I am not sure).

    Regarding the probability mass function, p(i), I have no idea how to calculate it.

    Any ideas would be much appreciated.
     
  2. jcsd
  3. Jul 1, 2009 #2

    EnumaElish

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    You need to find a function M such that M(n)(0)=2^n for n = 1, 2, 3, ... through infinity.
     
  4. Jul 2, 2009 #3
    Thanks for the help!

    Your answer would indicate that it isn't necessary to compute any derivatives to determine the function M. Is that correct?

    Also, do you have any idea how to go about computing the probability mass fctn p(i)?
     
  5. Jul 2, 2009 #4

    HallsofIvy

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    How does saying you need a function whos every derivative, at 0, is 2^n "indicate that it isn't necessary to compute any derivatives"?

    The "moment generating function" has the moments as coefficients in its McLaurin series expansion. That's why "2x" is not correct. Written as a McLaurin series, that is, of course, 0+ 2x+ 0x^2+ 0x^3+ ..., not [itex]1+ 2x+ 4x^3+ 8x^3+ ...= \sum_{n=0}^\infty 2^nx^n= \sum_{n=0}^\infty (2x)^n[/itex]. Can you identify the function given by that sum?
     
  6. Jul 2, 2009 #5

    Dick

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    I'm not at all an expert here, but I think Halls is missing an n! in the sum. Once you get the moment generating function then the probability distribution is a fourier transform of the characteristic function. Which appears, in stat land, to be the moment generating function with an i*t substituted for t. If you follow all of this and correct my mistakes (I just looked it up this afternoon) then you'll realize you actually could have guessed the PDF whose moments are 2^n. It's obvious in retrospect. Where are you, statdad and EnumaElish?
     
    Last edited: Jul 2, 2009
  7. Jul 3, 2009 #6
    EnumaElish, Halls (modulo the missing n!), and Dick are all right, of course.

    Here's my long way. If you use

    [tex]e^{z}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\dots[/tex]

    then you can make the formal derivation

    [tex]M(t)=E[tX]=\int e^{tx}f(x)\,dx=\intf(x)\,dx+t\int xf(x)\,dx+\frac{t^2}{2!}\int x^2f(x)\,dx+\frac{t^3}{3!}\int x^3f(x)\,dx+\dots[/tex]

    i.e.

    [tex]M(t)=1+tE[X]+\frac{t^2}{2!}E[X^2]+\frac{t^3}{3!}E[X^3]+\dots[/tex]


    Now substitute [tex]E[X^n]=2^n[/tex] to obtain M(t).

    Furthermore, then you can use the formula for e^z (backwards) to write M(t) in a very simple form.

    At this point, you should recognize that M(t) is the m.g.f. of a p.m.f. that has all its mass concentrated at one point.

    That is, p(x)=1 when x=c, where c is practically obvious.

    Finally, once you see what p(x) is, you realize you could have "obviously" jumped directly to the correct p(x) from the beginning, and then obtained M(t) from that, as everyone suggested.
     
  8. Jul 4, 2009 #7
    Thanks for the help (and patience), guys; I finally understand it.
     
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