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MGF of chi square

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data
    rvX has [tex] f(x) = \alpha \exp^{-\alpha x} , and \ W = 2n \alpha \overline {X}[/tex] defines a random sample from the distribution.
    Use moment generating function techniques to show that the distribution of W is chi-square on 2n degrees of freedom.
    2. Relevant equations
    3. The attempt at a solution
    Well......
    Ive let [tex] \alpha = \frac {1}{\beta}[/tex], then [tex]f(x)[/tex] ~ [tex] exp(\beta)[/tex]
    [tex]M_x(t) = (1 - \beta t)^{-1}[/tex]
    mgf of W with w~chisquare(2n)
    [tex] M_w(t) = (1 - 2t)^{-2v} [/tex]

    I don't really know what to do after this. Any help appreciated
     
  2. jcsd
  3. Sep 28, 2008 #2

    statdad

    User Avatar
    Homework Helper

    I'll use [tex] T [/tex] for the statistic of interest.

    [tex]
    T = 2n\alpha \overline X = 2 \alpha \sum_{i=1}^n X_i
    [/tex]

    You know the m.g.f. of each [tex] X_i [/tex] (they are iid). When you begin to calculate

    [tex]
    \int_{-\infty}^\infty e^{st} \, dt = E[e^{st}]
    [/tex]

    remember that [tex] t [/tex] is a sum, and use properties of exponents and expected values. You should wind up with a product that will lead you to the answer. (This all relies on the fact that the moment-generating function uniquely identifies the [tex] \chi^2 [/tex] distribution.)
     
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