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Math Help - Looking for a better method (convergence in mean square)

  1. #1
    Moo
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    Looking for a better method (convergence in mean square)

    Hi,

    I'm preparing a lesson with a student, and since he's not a major in maths, I'm looking for a better solution than mine, because it looks too "mathematical"

    We have a sample M=\max(|X_1|,\dots,|X_n|) (iid) where Xi follows a uniform distribution over [-\theta,\theta], where \theta>0 is a parameter.

    Previously, it was asked to find the mean and the variance of Xi. And the cdf of |Xi| (y/theta for y in [0,theta])

    Question 3) asks for the cdf of M, which is G(u)=\frac{u^n}{\theta^n}, for u\in[0,\theta]

    It's ok from here.

    Then, question 4) asks for the mean of M.
    What I did is taking the derivative of G, and then \mathbb{E}(M)=\int_0^\theta u G'(u) ~du=\dots=\frac{n}{n+1}\cdot \theta

    I'm already concerned that this is a too complicated method... So if you have a better one, denote it (1)

    Second part of question 4) asks k such that W=kM is an unbiased estimator for \theta
    Nothing magic here, k=\frac{n+1}{n}


    Third part of question 4) asks to show that W converges in mean square.
    It is not mentioned that it converges to \theta. So how can I explain to the boy that it should converge to theta ?
    This is the most awful part, because what I did is to use the property that :
    W converges to a constant c \Longleftrightarrow \lim_{n\to\infty}\mathbb{E}(W)=\theta and \lim_{n\to\infty} \mathbb{V}\text{ar}(W)=0

    But calculating the variance of W is a really ugly... So if you have a better solution, please denote it (2)

    And if you think these are the only ways to solve the questions, just eat a T-bone steak for dinner


    Thanks
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  2. #2
    MHF Contributor matheagle's Avatar
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    I'm not exactly sure what your asking, and I'm almost certainly sure you'll yell at me.

    BUT I would ignore chebyshev's and just show that W_n\to\theta as n goes to infinity

    That is, just use the definition E|W_n-\theta|^2\to 0

    And you do need to calculate the moments, but instead of the variance I would just use

    E|W_n-\theta|^2=E(W^2_n)-2\theta E(W_n)+\theta^2

    and show that this approaches zero as n goes to infinity.
    Last edited by matheagle; August 13th 2009 at 12:22 PM.
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  3. #3
    Moo
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    Thanks, it simplifies a bit to use E|W-theta|
    And no, I won't yell
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