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Convergence in Probability am I doing something wrong?

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

    Let [itex]\bar{X_n}[/itex] denote the mean of a random sample of size n from a distribution that has pdf [itex]f(x) = e^{-x}[/itex], [itex]0<x<\infty[/itex], zero elsewhere.

    a) Show that the mgf of [itex]Y_n=\sqrt{n}(\bar{X_n}-1)[/itex] is [itex]M_{Y_n}(t) = [e^{t/\sqrt{n}} - (t/\sqrt{n})e^{t/\sqrt{n}}]^{-n}[/itex], [itex]t < \sqrt{n}[/itex]


    b) Find the limiting distribution [itex]Y_n[/itex] as [itex]n \rightarrow \infty[/itex]




    2. Relevant equations



    3. The attempt at a solution

    a) We know that the pdf of [itex]\bar{X_n}[/itex] is [itex]\Gamma(n,1)[/itex]

    [itex]M_{Y_n}(t) = (\frac{1}{1-t})^n[/itex]

    [itex]M_{Y_n}(t) = E(e^{tY_n}) = E(e^{t\sqrt{n}(\bar{X_n}-1)} = e^{-t\sqrt{n}}E(e^{t\sqrt{n}\bar{X_n}})[/itex]

    MGF of [itex]\bar{X_n}[/itex] evaluated at [itex]t\sqrt{n}[/itex]

    [itex]=e^{-t\sqrt{n}}(\frac{1}{1-t\sqrt{n}})^n = ((e^{t/\sqrt{n}})(1-t\sqrt{n}))^{-n}[/itex]

    This does not match the answer they give because t is multiplied by [itex]\sqrt{n}[/itex] instead of being divided by it. But I just can't see where I went wrong...

    Thanks in advance
     
  2. jcsd
  3. Feb 2, 2013 #2
    The sum of n independent Exp(1) is Gamma(n,1), so if I'm not mistaken, the distribution of the mean is 1/n*Gamma(n,1) instead of Gamma(n,1).
     
  4. Feb 3, 2013 #3
    Thanks a lot. Can you explain a little more about why it needs to be [itex]\frac{1}{n}\Gamma(n,1)[/itex]?

    Is it because the sample mean is [itex]\frac{X_1 + X_2 + ... + X_n}{n}[/itex]? And since we know that the sum of n independent Exp(1) is Gamma(n,1), [itex]X_1 + ... + X_n = \Gamma(n,1)[/itex], right? Is that why we divide by n?
     
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