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Statistics - Moment Generating Functions

  1. Jun 30, 2013 #1
    1. The problem statement, all variables and given/known data

    The moment generating function (m.g.f.) of a random variable X is defined as the Expected value of etX:

    M(t) = E(etX).

    The series expansion of etX is:

    etX = 1 + tX + (t2X2)/(2!) + (t3X3)/(3!) + ...

    Hence,

    M(t) = E(etx) = 1 + tμ1' + (t2μ2')/2! + (t3μ3')/3! + ...


    If we differentiate M(t) r times with respect to t and then set t = 0 we shall therefore obtain the rth moment about the origin μr'.

    For example, the m.g.f. of the uniform distribution is:
    M(t) = E(etX) = [itex] \int_0^1 e^{tx} [/itex] dx = [itex] \dfrac{1}{t}e^{tx} [/itex] |[itex]_0^1 = \dfrac{1}{t}(e^t - 1) [/itex]

    = [itex] 1 + \dfrac{t}{2!} + \dfrac{t^2}{3!} + \dfrac{t^3}{4!} + [/itex] ...

    Differentiate this series r times and then set t = 0. Show that μr' = [itex] \dfrac{1}{r + 1} [/itex].

    2. Relevant equations



    3. The attempt at a solution

    I tried to do this by differentiating a few times to find a pattern:


    [itex] 1 + \dfrac{t}{2!} + \dfrac{t^2}{3!} + \dfrac{t^3}{4!} + \dfrac{t^4}{5!} [/itex] ...

    Differentiate once:
    [itex] \dfrac{1}{2!} + \dfrac{2t}{3!} + \dfrac{3t^2}{4!} + \dfrac{4t^3}{5!} [/itex] ...

    Differentiate twice:
    [itex] \dfrac{2}{3!} + \dfrac{6t}{4!} + \dfrac{12t^2}{5!} [/itex] ...

    Differentiate thrice:
    [itex] \dfrac{6}{4!} + \dfrac{24t}{5!} [/itex] ...

    Since t will be set to zero, only the first terms of each differentiation "survive:"

    [itex] \dfrac{1}{2!}, \dfrac{2}{3!}, \dfrac{6}{4!}, \dfrac{24}{5!}, ... [/itex]

    The pattern from this I noticed was: [itex] \dfrac{r!}{(r + 1)!} [/itex], where (r+1)! = (r+1)r!

    So, [itex] \dfrac{r!}{(r+1)r!} = \dfrac{1}{r+1} [/itex].

    This seems correct as I've obviously arrived at the correct expression, but I am wondering if I have approached this correctly and if there might have been any errors in my thought process.

    I am also a little bit lost conceptually as to why t is set equal to 0. What does setting t = 0 signify?

    I'm self-studying this topic and I'd really like to have a full understanding of what's going on rather than a superficial understanding where I'm just accepting that it says set t = 0. I'd appreciate any amount of elucidation!

    Thanks!
     
  2. jcsd
  3. Jul 1, 2013 #2

    BruceW

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    Homework Helper

    your method is good. And as for why t is set to zero... I just think of it as a useful 'trick'. I guess you can think of it like you are just considering the zeroth order term of the series. If you want to look into more detail, then maybe look at the wiki page on "characteristic function (probability theory)" Which I think is the nice generalisation of the generating function.
     
  4. Jul 1, 2013 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    You are trying to get the Maclaurin expansion, which reads as
    [tex] f(t) = f(0) + t f'(0) + \frac{t^2}{2!} f''(0) + \cdots + \frac{t^n}{n!} f^{(n)}(0) + \cdots [/tex]
    The coefficients are obtained by getting higher derivatives and evaluating them at ##t=0##.

    Another way to see this for your specific case is to look at the higher derivatives of the different terms. We have
    [tex] \left(\frac{d}{dt}\right)^r t^k = \left\{
    \begin{array}{cl} 0 &, \; r > k\\
    r! &, \; r = k \\
    r! t^{k-r}&, \; r < k
    \end{array}
    \right. [/tex]
    When we set t = 0 we are just picking out the rth term.
     
  5. Jul 5, 2013 #4
    That makes a lot of sense. I appreciate it both of you!
     
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