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Math Help - Shifted exponential

  1. #1
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    Shifted exponential

    Let X1, ..., Xn be iid r.v. with common pdf
    f(x) = e^-(x-θ) , x>θ , -inf < θ < inf
    = 0 elsewhere

    This pdf is called the shifted exponential. Let Yn = min{X1,...,Xn}. Prove that Yn --> θ in probability, by obtaining the cdf and the pdf of Yn.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I not only gave this lecture today I work with order stats.
    so far I have ...

    Here it is....
    Forget about pdf's, you can differentiate the cdf if you wish.
    F(x)=1-exp{-(x-theta)} if x exceeds theta and zero otherwise.

    So the cdf of Y_n is (I'm using < since it's continuous and I don't want to write <=)
    G(y)=P(Y_n<y) = 1-P(Y_n>y)=1-[1-F(y)]^n
    =1- exp{-n(x-theta)}
    THUS
    P(|Y_n-theta|>espislon)=P(theta-epsilon<Y_n<theta+epsilon)
    =P(theta<Y_n<theta+epsilon) due to the support
    =G(theta+epsilon)-G(theta)
    =(1- exp{-n(theta+epsilon-theta)})-(1- exp{-n(theta-theta)})
    =1-exp{-n epsilon} -1 +1
    =1-exp{-n epsilon}
    ->1
    for all epsilon as n-> infinity
    Last edited by mr fantastic; February 22nd 2009 at 01:07 AM. Reason: Merged posts
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