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Probability: How to prove a function is m.g.f.?

  1. Feb 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Question : Prove [itex] \varphi (t) = \sum_{i=0}^{ \infty } a_icos(it) [/itex] is a moment generating function (m.g.f.) and determine its corresponding probability density function (p.d.f.)
    when [itex] \sum_{i=0}^{ \infty } a_i=1 [/itex] holds for [itex] a_i \geq 0 [/itex].

    2. Relevant equations

    Nothing special.

    3. The attempt at a solution

    I really don't know what to do with this question, all I know is [itex] \varphi (0) = \sum_i p(x_i) = 1 [/itex] in the discrete case.

    For the current one,

    [itex] \varphi (0) = \sum_{i=0}^{ \infty } a_i = 1 [/itex]

    This is a proof? It cannot be so easy, and how to determine the p.d.f. when its m.g.f. is given?

    Thank you in advance!
     
  2. jcsd
  3. Feb 25, 2014 #2

    haruspex

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    It says "a" m.g.f., so I presume we get to choose whether it's discrete or continuous. Let's try discrete.
    The given definition of φ(t) cannot be right since it gives a negative second moment. I suspect it should say "characteristic function". This fits with φ(t) being the usual choice for representing a characteristic function, whereas M(t) is standard for an m.g.f. See http://en.wikipedia.org/wiki/Moment-generating_function.
    The choice of i as the index may lead to some confusion. Let's use r instead.
    Putting that aside for now, can you post equations for the nth moment:
    1. In terms of the m.g.f. Treat n even and n odd separately.
    2. In terms of the p.d.f. (pr). Careful with the range for the sum here.
     
  4. Feb 25, 2014 #3

    Ray Vickson

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    Are you sure you copied the problem correctly? If you had "cosh" instead of "cos" there would be a fairly easy solution to the problem of finding a real random variable X with that mgf. However, as haruspex has pointed out the function as written cannot be the mgf (of a real random variable); it could be the mgf of a complex-valued random variable taking values along the imaginary axis.
     
  5. Feb 25, 2014 #4
    1. In terms of the m.g.f. Treat n even and n odd separately.

    Because

    [itex] \begin{cases} \varphi' (t) = - \sum_{r=0}^{ \infty }a_r r sin(rt)
    \\ \varphi'' (t) =- \sum_{r=0}^{ \infty }a_r r^2 cos(rt)
    \\ \varphi^{(3)} (t) = \sum_{r=0}^{ \infty }a_r r^3 sin(rt)
    \\ \varphi^{(4)} (t) = \sum_{r=0}^{ \infty }a_r r^4 cos(rt)
    \end{cases} [/itex]

    Then it repeats except the power of r, hence

    [itex] \varphi^{(n)} (t) = \begin{cases} (-1)^{g(n)} \sum_{r=0}^{ \infty }a_r r^n sin(rt) & n \;\; is \;\; odd
    \\ (-1)^{g(n)} \sum_{r=0}^{ \infty }a_r r^n cos(rt) & n \;\; is \;\; even
    \end{cases} [/itex]

    where

    [itex] g(n) =\begin{cases}1 & if \;\; n\;\; mod\;\; 4 = 1\;\; or\;\; 2
    \\0 & if \;\; n\;\; mod\;\; 4 = 3\;\; or\;\; 0
    \end{cases} [/itex]

    Hence

    [itex] \varphi^{(n)} (0) = \begin{cases} 0 & n \;\; is \;\; odd
    \\ (-1)^{g(n)} \sum_{r=0}^{ \infty }a_r r^n & n \;\; is \;\; even
    \end{cases} [/itex]

    Is this correct?

    2. In terms of the p.d.f. (pr).

    [itex] E[x^n] = \sum_{x=0}^{ \infty }x^np(x) [/itex]

    Then I don’t know what to do next. I tried to compare this with the [itex] \varphi^{(n)} (0) [/itex], but found the minus prevents the [itex] p(x) [/itex] to have a consistent form, what shall I do?
     
  6. Feb 25, 2014 #5
    The problem statement is correct, "cos" has no "h" attached.

    Except the statement characteristic function is replaced by m.g.f., because I thought they are same in probability.

    Since no textbook on my hands mentioned characteristic function, it's my mistake to change the term.
     
  7. Feb 25, 2014 #6

    Ray Vickson

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    Mgfs and characteristic functions are not the same. For a real random variable ##X## the mgf is ##\phi(k) = E \exp(kX)## while the characteristic function is ##\chi(k) = E \exp(i k X)##. In other words, for real ##k## we have ##\chi(k) = \phi(i k)##; here, ##i = \sqrt{-1}##, not a summation index.

    You don't need a textbook on hand; Google is your friend.
     
  8. Feb 25, 2014 #7

    haruspex

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    Having sorted out the confusion between m.g.f and characteristic function, what does the above change to?
    You mean
    [itex] E[X^n] = \sum_{x=0}^{ \infty }x^np(x) [/itex]
    I said to be careful about the index range here!
    What is E[X], as determined by the characteristic function? What does that suggest about the range of x values?
     
  9. Feb 25, 2014 #8
    [itex] \varphi^{(n)} (t) = \begin{cases} -i \sum_{r=0}^{ \infty }a_r r^n sin(irt) & n \;\; is \;\; odd
    \\ \sum_{r=0}^{ \infty }a_r r^n cos(irt) & n \;\; is \;\; even
    \end{cases} [/itex]

    [itex] \varphi^{(n)} (0) = \begin{cases} 0 & n \;\; is \;\; odd
    \\ \sum_{r=0}^{ \infty }a_r r^n & n \;\; is \;\; even
    \end{cases} [/itex]

    Since [itex]E(X) = \varphi' (0) = 0 [/itex], X should have a symmetric distribution?

    Hence [itex] X \in (- \infty ,+ \infty ) [/itex], then [itex] p(x) = \frac{a_{|x|}}{2} [/itex]

    Is this correct?
     
  10. Feb 26, 2014 #9

    Ray Vickson

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    Are you saying ##P(X=0) = a_0##, ##P( X = 1) = P(X = -1) = \frac{1}{2}a_1 ##, ##P(X = 2) = P(X = -2) = \frac{1}{2} a_2,## etc? That would be a correct way of saying it.
     
  11. Feb 26, 2014 #10
    Yes, that is exactly what I mean!

    Thank you so much, Ray!

    And I still don't know how to prove a function is a characteristic function? Can you tell me how to do that?
     
  12. Feb 26, 2014 #11

    Ray Vickson

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    Basically, there is nothing to prove. You have constructed a random variable ##X##, and its characteristic function is the function you were given. End of story.
     
  13. Feb 26, 2014 #12
    Ray, thank you for your reply.

    A loop arises and I didn't notice it.
     
    Last edited: Feb 26, 2014
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