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Math Help - Convergence in probability

  1. #1
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    Convergence in probability (Uniform Distribution)

    let X_1,...,X_n are iid U(0, \theta). I am trying to show that \sqrt{T_n} \rightarrow^{p} \sqrt{\theta}(converges in probability to), where T_n is the largest\;order\; statistic

    what I did was using the pdf of max order statistic find

    E(T_n)\;=\;\dfrac{n\theta}{n+1}

    E({T_n}^2)=\dfrac{n\theta^2}{n+2}

    so, E\{(T_{n}-\theta)^2\} = \dfrac{2{\theta}^2}{(n+1)(n+2)}, which converges to 0, as n goes to infinity,

    so we can conclude by weak law of large numbers that T_n \rightarrow^{p} \theta \;as\; n \to \infty

    and this \sqrt{T_n} \rightarrow^{p} \sqrt{\theta} follows from the theorem of continuous mapping....

    is my approach to the proof correct?

    thank you.
    Last edited by chutiya; January 30th 2011 at 02:48 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I didn't check your calculations, but by Markov all you need is

    P\left(|T_n-\theta|\ge \epsilon\right)\le {E(T_n-\theta)^2\over \epsilon^2}\to 0

    Then you can use the fact that the square root is a continuous function giving

    \sqrt{T_n}\to \sqrt{\theta}

    So, you may need to add and subtract your mean in that second moment and expand
    And this is not the Weak Law of Large Numbers, this is not a sum.

    Like  E(T_n-\theta)^2=E[(T_n- {n\theta\over n+1}) +({n\theta\over n+1}-\theta)]^2
    Last edited by matheagle; January 30th 2011 at 07:41 PM.
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  3. #3
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    thank you matheagle..

    actually what I have done is:

    P(|T_n-\theta|\geq \epsilon)=P(|T_n - \theta|^2 \geq \epsilon^2) \leq \dfrac{E[(T_n-\theta)]^2}{\epsilon^2}=\dfrac{2\theta^2}{\epsilon^2(n+1)(  n+2)} which goes to 0 as n goes to infty
    Last edited by chutiya; January 30th 2011 at 07:00 PM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    The idea is right, but I didn't check your calculations. I do see a power missing on epsilon though.
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  5. #5
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    Thank you. I was typing too fast i guess.. thats an epsilon squared in the den..

    there's one more question about convergence which I wanted to ask..

    if X_n is a Poisson(n) RV with positive integer rate parameter, show that as\; n\to \infty, \dfrac{X_n-E(X_n)}{\sqrt{Var(X_n)}}} \rightarrow^d \; N(0,1)

    I think this involves CLT, but I am confused on how to solve this.. could you help?

    thanks.
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  6. #6
    MHF Contributor matheagle's Avatar
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    Use the fact that the sum of i.i.d. Poisson's with mean one is a Poisson with mean n.
    NOW use the classical CLT on the sum.
    Just multiply the sample mean by n to get a CLT for the sum.

     {\bar X_n- E(\bar X_n)\over \sqrt{V(\bar  X_n)}} \to N(0,1)

    {\bar X_n- E(\bar X_n)\over \sqrt{V(\bar X_n)}} ={n\bar X_n- nE(\bar X_n)\over n\sqrt{V(\bar X_n)}}

    This n\bar X_n=\sum_{i=1}^nY_i where each Y_i is a Poisson(1)
    Last edited by matheagle; January 30th 2011 at 07:42 PM.
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