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Thread: Convergence of uniform, independent RV's

  1. #1
    Member Maccaman's Avatar
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    Convergence of uniform, independent RV's

    $\displaystyle X_1, X_2,..... $ are independent random variables that all have a uniform distribution, U(0,1).

    Prove that if $\displaystyle M_n = max (X_1, X_2, ..., X_n) $ then as n approaches infinity, $\displaystyle n(1-M_n) $ converges in distribution to an Exp(1) random variable.

    I've done a little bit but I have confused myself and I don't know what to do with it now (and im not even sure if Im on the right track ).

    So

    $\displaystyle Pr(n(1-M_n) \leq x) = Pr(1 - \frac{x}{n} \leq M_n)$

    $\displaystyle = 1 - Pr(M_n < 1 - \frac{x}{n})$

    $\displaystyle = 1 - Pr(\forall n, X_n < 1 - \frac{x}{n}) $

    $\displaystyle = 1 - \prod_{i=1}^n \ Pr(X_i < 1 - \frac{x}{n})$

    $\displaystyle = 1 - (1-\frac{x}{n})^n$

    and this is where I get lost. I think maybe.....

    $\displaystyle = 1 - exp(n \ln (1 - \frac{x}{n}))$

    but then where to go from here I dont know, which makes me doubt the accuracy of what I have done. Thanks for any help you can provide me with.
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  2. #2
    MHF Contributor matheagle's Avatar
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    you did the hard work, you just need to review calc 2

    $\displaystyle \biggl(1-\frac{x}{n}\biggr)^n\to e^{-x}$ done
    Last edited by matheagle; Aug 22nd 2009 at 09:55 PM.
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  3. #3
    Member Maccaman's Avatar
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    Ahhh, off course. Why I overcomplicate things sometimes I'll never know
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  4. #4
    MHF Contributor matheagle's Avatar
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    Don't be so hard on yourself.
    We have Moo at this forum for that.
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