Moment Generating Function (proof of definition)

  1. 1. The problem statement, all variables and given/known data
    Prove that for a random variable [tex]X[/tex] with continuous probability distribution function [tex]f_X(x)[/tex] that the Moment Generating Function, defined as

    [tex]
    M_X(t) := E[e^{tX}]
    [/tex]

    is

    [tex]
    M_X(t) = \int_x^{\infty}e^{tx}f_X(x)dx
    [/tex]

    2. Relevant equations

    Above and

    [tex]
    E[X] = \int_{-\infty}^{\infty}xf_X(x)dx
    [/tex]

    3. The attempt at a solution

    This expression is given in so many textbooks and the ones that I have read all skip over this derivation. I want to be able to prove (1) to myself.

    Proof:
    Write the exponential function as a Maclaurin series:

    [tex]
    M_X(t) = E[e^{tX}]
    [/tex]

    [tex]
    = E[1+tX+\frac{t^2}{2!}X^2+\frac{t^3}{3!}X^3+...]
    [/tex]

    Since [tex]E[1] = 1[/tex] and the [tex]E[t^n/n!]=t^n/n![/tex] because they are constant and the expectation of a constant is itself you get:

    [tex]
    = 1+tE[X]+\frac{t^2}{2!}E[X^2]+\frac{t^3}{3!}E[X^3]+...
    [/tex]


    ...also using the linearity of E. Now, writing the series as a sum:

    [tex]
    =\sum_{t=0}^{\infty}\frac{t^n}{n!}E[X^n]
    [/tex]

    And extracting the exponential:

    [tex]
    =e^t\sum_{n=0}^{\infty}E[X^n]
    [/tex]

    Now I am stuck! I know that I am meant to use

    [tex]
    E[X] = \int_{-\infty}^{\infty}xf_X(x)dx
    [/tex]

    but I have [tex]E[X^n][/tex] and I also have [tex]e^t[/tex] and not [tex]e^{tx}[/tex].
     
    Last edited: Mar 24, 2011
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,960
    Staff Emeritus
    Science Advisor

    There isn't really much to prove. For any continuous probability distribution with density function f(x), the Expected value of any function u(x) is defined to be
    [tex]E(u(x))= \int_{-\infty}^\infty u(x)f(x)dx[/itex]

    Replace [itex]u(x)[/itex] with [itex]e^{tx}[/itex] and you have it. It is true that the whole point of the "moment generating function" is that the coefficients of the powers of x in a power series expansion are the "moments" of the probability distribution, but that doesn't seem to me to be relevant to this question. I see no reason to write its Taylor series.
     
  4. Good, okay that makes sense.

    Then I suppose all I had to do was prove

    [tex]
    E(u(x))= \int_{-\infty}^\infty u(x)f(x)dx
    [/tex]

    and then substitute [tex]u(x)[/tex] with [tex]e^{tx}[/tex] as you said and I'm done.

    But once again there is nothing to prove because it is a definition.
     
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