Order statistic of i.i.d exponential($lambda$) random variables, $X_{(n, k_n)}$ convergence in probability












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Suppose that $X_1,X_2$,....are iid from exponential($lambda$).For n $geq$ 1, let $X_{(n,1)}le X_{(n,2)}le X_{(n,3)}le.......le X_{(n,n)}$ be the order statistics of $X_1,X_2....X_n$. Suppose that $k_n$ is a sequence of integers satisfying $1le k_n le n$ for all n, and


${lim_{xto infty}frac{k_n}{n}} = pin(0,1)$


Show that as n$toinfty$



$qquadqquadqquadqquadmathit{{X_{(n, k_n)}to-frac{1}{lambda}log(1-p)}}$


I am thankful for any help.






probability convergence order-statistics exponential-distribution







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  • $begingroup$
    Why the crazy encoding?
    $endgroup$
    – Did
    Dec 3 '18 at 7:04










  • $begingroup$
    was my first time thats why
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:08
















0












$begingroup$


Suppose that $X_1,X_2$,....are iid from exponential($lambda$).For n $geq$ 1, let $X_{(n,1)}le X_{(n,2)}le X_{(n,3)}le.......le X_{(n,n)}$ be the order statistics of $X_1,X_2....X_n$. Suppose that $k_n$ is a sequence of integers satisfying $1le k_n le n$ for all n, and


${lim_{xto infty}frac{k_n}{n}} = pin(0,1)$


Show that as n$toinfty$



$qquadqquadqquadqquadmathit{{X_{(n, k_n)}to-frac{1}{lambda}log(1-p)}}$


I am thankful for any help.






probability convergence order-statistics exponential-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why the crazy encoding?
    $endgroup$
    – Did
    Dec 3 '18 at 7:04










  • $begingroup$
    was my first time thats why
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:08














0












0








0





$begingroup$


Suppose that $X_1,X_2$,....are iid from exponential($lambda$).For n $geq$ 1, let $X_{(n,1)}le X_{(n,2)}le X_{(n,3)}le.......le X_{(n,n)}$ be the order statistics of $X_1,X_2....X_n$. Suppose that $k_n$ is a sequence of integers satisfying $1le k_n le n$ for all n, and


${lim_{xto infty}frac{k_n}{n}} = pin(0,1)$


Show that as n$toinfty$



$qquadqquadqquadqquadmathit{{X_{(n, k_n)}to-frac{1}{lambda}log(1-p)}}$


I am thankful for any help.






probability convergence order-statistics exponential-distribution







share|cite|improve this question











$endgroup$




Suppose that $X_1,X_2$,....are iid from exponential($lambda$).For n $geq$ 1, let $X_{(n,1)}le X_{(n,2)}le X_{(n,3)}le.......le X_{(n,n)}$ be the order statistics of $X_1,X_2....X_n$. Suppose that $k_n$ is a sequence of integers satisfying $1le k_n le n$ for all n, and


${lim_{xto infty}frac{k_n}{n}} = pin(0,1)$


Show that as n$toinfty$



$qquadqquadqquadqquadmathit{{X_{(n, k_n)}to-frac{1}{lambda}log(1-p)}}$


I am thankful for any help.






probability convergence order-statistics exponential-distribution




probability convergence order-statistics exponential-distribution






share|cite|improve this question















share|cite|improve this question













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edited Dec 3 '18 at 7:03









Did

247k23222458




247k23222458










asked Dec 3 '18 at 4:19









AmeliaAmelia

329




329












  • $begingroup$
    Why the crazy encoding?
    $endgroup$
    – Did
    Dec 3 '18 at 7:04










  • $begingroup$
    was my first time thats why
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:08


















  • $begingroup$
    Why the crazy encoding?
    $endgroup$
    – Did
    Dec 3 '18 at 7:04










  • $begingroup$
    was my first time thats why
    $endgroup$
    – Amelia
    Dec 3 '18 at 7:08
















$begingroup$
Why the crazy encoding?
$endgroup$
– Did
Dec 3 '18 at 7:04




$begingroup$
Why the crazy encoding?
$endgroup$
– Did
Dec 3 '18 at 7:04












$begingroup$
was my first time thats why
$endgroup$
– Amelia
Dec 3 '18 at 7:08




$begingroup$
was my first time thats why
$endgroup$
– Amelia
Dec 3 '18 at 7:08










1 Answer
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$begingroup$

Fix $epsilon > 0$.



Let $Y_1 sim text{Binom}(n, (1-p) e^{lambda epsilon})$
and $Y_2 sim text{Binom}(n, (1-p) e^{-lambda epsilon})$.
begin{align}
P(|X_{n,k_n} + log(1-p)/lambda| > epsilon)
&le P(X_{n,k_n} < - log(1-p)/lambda - epsilon)
+ P(X_{n,k_n} > - log(1-p)/lambda + epsilon)
\
&= P(text{less than $n-k_n$ of the $X_i$ are $ge - log(1-p)/lambda - epsilon$})
+ P(text{more than $n-k_n$ of the $X_i$ are $ge - log(1-p)/lambda + epsilon$})
\
&= P(Y_1 < n-k_n) + P(Y_2 > n-k_n)
\
&= P(Y_1/n < 1 - k_n/n) + P(Y_2/n > 1 - k_n/n).
end{align}
By the law of large numbers, $Y_1/n to (1-p) e^{lambda epsilon} > 1 - p$ in probability, so the first term will tend to zero. Similarly, the second term will tend to zero.






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    1 Answer
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    1 Answer
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    active

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    1












    $begingroup$

    Fix $epsilon > 0$.



    Let $Y_1 sim text{Binom}(n, (1-p) e^{lambda epsilon})$
    and $Y_2 sim text{Binom}(n, (1-p) e^{-lambda epsilon})$.
    begin{align}
    P(|X_{n,k_n} + log(1-p)/lambda| > epsilon)
    &le P(X_{n,k_n} < - log(1-p)/lambda - epsilon)
    + P(X_{n,k_n} > - log(1-p)/lambda + epsilon)
    \
    &= P(text{less than $n-k_n$ of the $X_i$ are $ge - log(1-p)/lambda - epsilon$})
    + P(text{more than $n-k_n$ of the $X_i$ are $ge - log(1-p)/lambda + epsilon$})
    \
    &= P(Y_1 < n-k_n) + P(Y_2 > n-k_n)
    \
    &= P(Y_1/n < 1 - k_n/n) + P(Y_2/n > 1 - k_n/n).
    end{align}
    By the law of large numbers, $Y_1/n to (1-p) e^{lambda epsilon} > 1 - p$ in probability, so the first term will tend to zero. Similarly, the second term will tend to zero.






    share|cite|improve this answer









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      1












      $begingroup$

      Fix $epsilon > 0$.



      Let $Y_1 sim text{Binom}(n, (1-p) e^{lambda epsilon})$
      and $Y_2 sim text{Binom}(n, (1-p) e^{-lambda epsilon})$.
      begin{align}
      P(|X_{n,k_n} + log(1-p)/lambda| > epsilon)
      &le P(X_{n,k_n} < - log(1-p)/lambda - epsilon)
      + P(X_{n,k_n} > - log(1-p)/lambda + epsilon)
      \
      &= P(text{less than $n-k_n$ of the $X_i$ are $ge - log(1-p)/lambda - epsilon$})
      + P(text{more than $n-k_n$ of the $X_i$ are $ge - log(1-p)/lambda + epsilon$})
      \
      &= P(Y_1 < n-k_n) + P(Y_2 > n-k_n)
      \
      &= P(Y_1/n < 1 - k_n/n) + P(Y_2/n > 1 - k_n/n).
      end{align}
      By the law of large numbers, $Y_1/n to (1-p) e^{lambda epsilon} > 1 - p$ in probability, so the first term will tend to zero. Similarly, the second term will tend to zero.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Fix $epsilon > 0$.



        Let $Y_1 sim text{Binom}(n, (1-p) e^{lambda epsilon})$
        and $Y_2 sim text{Binom}(n, (1-p) e^{-lambda epsilon})$.
        begin{align}
        P(|X_{n,k_n} + log(1-p)/lambda| > epsilon)
        &le P(X_{n,k_n} < - log(1-p)/lambda - epsilon)
        + P(X_{n,k_n} > - log(1-p)/lambda + epsilon)
        \
        &= P(text{less than $n-k_n$ of the $X_i$ are $ge - log(1-p)/lambda - epsilon$})
        + P(text{more than $n-k_n$ of the $X_i$ are $ge - log(1-p)/lambda + epsilon$})
        \
        &= P(Y_1 < n-k_n) + P(Y_2 > n-k_n)
        \
        &= P(Y_1/n < 1 - k_n/n) + P(Y_2/n > 1 - k_n/n).
        end{align}
        By the law of large numbers, $Y_1/n to (1-p) e^{lambda epsilon} > 1 - p$ in probability, so the first term will tend to zero. Similarly, the second term will tend to zero.






        share|cite|improve this answer









        $endgroup$



        Fix $epsilon > 0$.



        Let $Y_1 sim text{Binom}(n, (1-p) e^{lambda epsilon})$
        and $Y_2 sim text{Binom}(n, (1-p) e^{-lambda epsilon})$.
        begin{align}
        P(|X_{n,k_n} + log(1-p)/lambda| > epsilon)
        &le P(X_{n,k_n} < - log(1-p)/lambda - epsilon)
        + P(X_{n,k_n} > - log(1-p)/lambda + epsilon)
        \
        &= P(text{less than $n-k_n$ of the $X_i$ are $ge - log(1-p)/lambda - epsilon$})
        + P(text{more than $n-k_n$ of the $X_i$ are $ge - log(1-p)/lambda + epsilon$})
        \
        &= P(Y_1 < n-k_n) + P(Y_2 > n-k_n)
        \
        &= P(Y_1/n < 1 - k_n/n) + P(Y_2/n > 1 - k_n/n).
        end{align}
        By the law of large numbers, $Y_1/n to (1-p) e^{lambda epsilon} > 1 - p$ in probability, so the first term will tend to zero. Similarly, the second term will tend to zero.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 5:00









        angryavianangryavian

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