1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook