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Math Help - Convergence in Distribution

  1. #1
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    Convergence in Distribution

    I am wondering if someone can give me some help on this problem.

    Let X_1, X_2, ... be i.i.d. and X_{(n)} = max_{1 \le i \le n}X_i. If X_i ~ Beta (1, \beta), find a value v so that
    n^v(1 - X_{(n)}) converges in distribution.

    I think it has to do with order statistics but I could be comletely off the right track.

    Thanks for your time!
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  2. #2
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    Quote Originally Posted by chabmgph View Post
    I am wondering if someone can give me some help on this problem.

    Let X_1, X_2, ... be i.i.d. and X_{(n)} = max_{1 \le i \le n}X_i. If X_i ~ Beta (1, \beta), find a value v so that
    n^v(1 - X_{(n)}) converges in distribution.

    I think it has to do with order statistics but I could be comletely off the right track.

    Thanks for your time!
    Start by getting the distribution of X_{(n)}.
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  3. #3
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    Ok. Let see if I can at least get this right...

    f_{X_i}=(1-x)^{\beta-1} since X_i \sim Beta(1, \beta)

    Then F_{X_i}=\int^x_0 (1-t)^{\beta-1} dt= \frac{1}{\beta}(1-(1-x)^{\beta})
    And since X_n is the max, the p.d.f for it is
    f(x_n)=n*f(x_n)*F^{n-1}(x_n)
    =n*(1-x_n)^{\beta-1}*\frac{1}{\beta^{n-1}}(1-(1-x_n)^{\beta})^{n-1}

    Am I in the right track?

    Thanks again!
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  4. #4
    MHF Contributor matheagle's Avatar
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    That's not a valid density.
    You're leaving out the constant terms in your Beta.
    And, yes the max is the largest order statistic.
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  5. #5
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    I see where I messed up now...'cause I can't add...

    So instead, f_{X_i}=\beta(1-x)^{\beta-1}

    Then it will be F_{X_i}=\beta \int^x_0 (1-t)^{\beta-1} dt= 1-(1-x)^{\beta}
    And p.d.f will be
    f(x_n)=n*f(x_n)*F^{n-1}(x_n)
    =n\beta(1-x_n)^{\beta-1}(1-(1-x_n)^{\beta})^{n-1}

    Does that look better?

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  6. #6
    MHF Contributor matheagle's Avatar
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    Instead of doing all of that, why don't you get the CDF of the max directly?

    P(X_{(n)}\le a)=P(X_1\le a,...., X_n\le a)=F^n(a)

    Then let Y_n=n^v(1-X_{(n)}) and find its CDF.
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  7. #7
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    Quote Originally Posted by matheagle View Post
    Instead of doing all of that, why don't you get the CDF of the max directly?

    P(X_{(n)}\le a)=P(X_1\le a,...., X_n\le a)=F^n(a)
    Oh, right, since they are i.i.d, I guess I could have done that. >"<

    So the c.d.f is F(x_{(n)})=[1-(1-x_{(n)})^{\beta}]^n.

    Edit:And I just saw the last line of your hint, let me try to see if I can get it out. Puzzle time! lol
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  8. #8
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    So let Y_n=n^v(1-X_{(n)})

    Then F(Y_n)=P(Y_n\le y)
    =P(n^v(1-X_{(n)})\le y)
    =P(X_{(n)}\ge 1-\frac{y}{n^v})
    =1- P(X_{(n)}\le 1-\frac{y}{n^v})
    =1-F(1-\frac{y}{n^v})
    =1-[1-(1-(1-\frac{y}{n^v}))^{\beta}]^n
    =1-[1-(\frac{y}{n^v})^{\beta}]^n

    So now I have to prove that this converges in distribution.
    So as n approaches infinity, doesn't (\frac{y}{n^v})^{\beta} approaches zero?
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  9. #9
    MHF Contributor matheagle's Avatar
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    USE calc 2

    \bigg(1-{a\over x}\biggr)^x \to?????

    And you left out a power of n.
    Last edited by matheagle; December 28th 2009 at 11:00 PM.
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  10. #10
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    Quote Originally Posted by matheagle View Post
    USE calc 2

    \bigg(1-{a\over x}\biggr)^x \to?????

    And you left out a power of n.
    e^a!!!

    So v=\frac{1}{\beta} must be true?

    Then F(Y_n) \to 1-e^{y^{\beta}}

    And wow on the publication.

    By the way, you are only 5 hours from me. lol
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  11. #11
    MHF Contributor matheagle's Avatar
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    No problem, any time
    Last edited by matheagle; December 28th 2009 at 11:00 PM.
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  12. #12
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    Thank you again for your help!
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  13. #13
    MHF Contributor matheagle's Avatar
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    Nope...........that limit is e^{-a}
    you need a negative in the exponent

    Quote Originally Posted by chabmgph View Post
    e^{-a}!!!

    So v=\frac{1}{\beta} must be true?

    Then F(Y_n) \to 1-e^{-y^{\beta}}

    And wow on the publication.

    By the way, you are only 5 hours from me. lol
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  14. #14
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    Can't believe it..... I messed it up again.

    Thanks again for catching that!
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  15. #15
    MHF Contributor matheagle's Avatar
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    I noticed it the other night.
    I was hoping you or someone else would point it out.
    The exponent must be negative if you want F(y) to approach 1 as y goes to infinity.
    And there is a power of n still missing.
    Last edited by matheagle; December 19th 2009 at 11:51 PM.
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