Chernoff Bound Variantion.











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How can we prove
$latex P( X geq (1+delta)mu) leq {e}^{frac{-delta}{3}{mu}} quad where quad delta > 1 $ -----------(1)



When I have already proved that:
$latex ( X geq (1+delta)mu) leq {e}^{frac{-delta^2}{3}{mu}} quad where quad 0 leq delta leq 1 $ -----------(2)



How can i extend the second Prof to show that (1) is also true.




Solution to 2



$ P( X geq (1+delta)mu) leq [frac {e^{delta}} {(1 + delta)^{1 +
> delta}}]^{mu} leq {e}^{frac{-delta^2}{3}{mu}} quad where quad
> 0 leq delta leq 1 quad quad $ (1)



Consider the denominator part to be,



$ {(1 + delta)^{1 + delta}} = e^{(1+delta)ln(1+delta)} quad
> quad $(2)



We can apply Taylor’s Series on,



$ {(1+delta)ln(1+delta)} quad As quad delta=[0,1] $



Therefore,



$ ln(1+delta) = delta - frac{delta^2}{2} + frac{delta^3}{3} -
> frac{delta^4}{4} + ... quad quad $(a)



$ delta ln(1+delta) = delta^2 - frac{delta^3}{2} +
> frac{delta^4}{3} - frac{delta^5}{4} + ... quad quad $ (b)



Adding Eq. a and b and taking $ ln(1+delta) $ common, we get,



$ (1+delta) ln(1+delta) = delta + frac{delta^2}{2} -
> frac{delta^3}{6} + ... $



Ignoring higher order terms,



$ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
> frac{delta^3}{6} $



We can further write it as,



$ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
> frac{delta.delta^2}{6} $



And take an assumption such that,



$ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
> frac{delta^2}{6} $



$ (1+delta) ln(1+delta) geq delta + frac{delta^2}{3} $



Now, from Eq. 2,



$ {(1 + delta)^{1 + delta}} = e^{(1+delta)ln(1+delta)} =
> e^{delta+ frac{delta^2}{3}} $



Therefore, Eq. 1 becomes,



$ P( X geq (1+delta)mu) leq [frac {e^{delta}} {(1 + delta)^{1 +
> delta}}]^{mu}= [frac {e^{delta}} {e^{delta+
> frac{delta^2}{3}}}]^{mu} = [ {e^{delta - delta-
> frac{delta^2}{3}}}]^{mu} $



$ P( X geq (1+delta)mu) leq {e^{-frac{delta^2}{3}}}^{mu} $







probability-theory inequality







share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    How can we prove
    $latex P( X geq (1+delta)mu) leq {e}^{frac{-delta}{3}{mu}} quad where quad delta > 1 $ -----------(1)



    When I have already proved that:
    $latex ( X geq (1+delta)mu) leq {e}^{frac{-delta^2}{3}{mu}} quad where quad 0 leq delta leq 1 $ -----------(2)



    How can i extend the second Prof to show that (1) is also true.




    Solution to 2



    $ P( X geq (1+delta)mu) leq [frac {e^{delta}} {(1 + delta)^{1 +
    > delta}}]^{mu} leq {e}^{frac{-delta^2}{3}{mu}} quad where quad
    > 0 leq delta leq 1 quad quad $     (1)



    Consider the denominator part to be,



    $ {(1 + delta)^{1 + delta}} = e^{(1+delta)ln(1+delta)} quad
    > quad $    (2)



    We can apply Taylor’s Series on,



    $ {(1+delta)ln(1+delta)} quad As quad delta=[0,1] $



    Therefore,



    $ ln(1+delta) = delta - frac{delta^2}{2} + frac{delta^3}{3} -
    > frac{delta^4}{4} + ... quad quad $    (a)



    $ delta ln(1+delta) = delta^2 - frac{delta^3}{2} +
    > frac{delta^4}{3} - frac{delta^5}{4} + ... quad quad $     (b)



    Adding Eq. a and b and taking $ ln(1+delta) $ common, we get,



    $ (1+delta) ln(1+delta) = delta + frac{delta^2}{2} -
    > frac{delta^3}{6} + ... $



    Ignoring higher order terms,



    $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
    > frac{delta^3}{6} $



    We can further write it as,



    $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
    > frac{delta.delta^2}{6} $



    And take an assumption such that,



    $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
    > frac{delta^2}{6} $



    $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{3} $



    Now, from Eq. 2,



    $ {(1 + delta)^{1 + delta}} = e^{(1+delta)ln(1+delta)} =
    > e^{delta+ frac{delta^2}{3}} $



    Therefore, Eq. 1 becomes,



    $ P( X geq (1+delta)mu) leq [frac {e^{delta}} {(1 + delta)^{1 +
    > delta}}]^{mu}= [frac {e^{delta}} {e^{delta+
    > frac{delta^2}{3}}}]^{mu} = [ {e^{delta - delta-
    > frac{delta^2}{3}}}]^{mu} $



    $ P( X geq (1+delta)mu) leq {e^{-frac{delta^2}{3}}}^{mu} $







    probability-theory inequality







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      How can we prove
      $latex P( X geq (1+delta)mu) leq {e}^{frac{-delta}{3}{mu}} quad where quad delta > 1 $ -----------(1)



      When I have already proved that:
      $latex ( X geq (1+delta)mu) leq {e}^{frac{-delta^2}{3}{mu}} quad where quad 0 leq delta leq 1 $ -----------(2)



      How can i extend the second Prof to show that (1) is also true.




      Solution to 2



      $ P( X geq (1+delta)mu) leq [frac {e^{delta}} {(1 + delta)^{1 +
      > delta}}]^{mu} leq {e}^{frac{-delta^2}{3}{mu}} quad where quad
      > 0 leq delta leq 1 quad quad $       (1)



      Consider the denominator part to be,



      $ {(1 + delta)^{1 + delta}} = e^{(1+delta)ln(1+delta)} quad
      > quad $      (2)



      We can apply Taylor’s Series on,



      $ {(1+delta)ln(1+delta)} quad As quad delta=[0,1] $



      Therefore,



      $ ln(1+delta) = delta - frac{delta^2}{2} + frac{delta^3}{3} -
      > frac{delta^4}{4} + ... quad quad $      (a)



      $ delta ln(1+delta) = delta^2 - frac{delta^3}{2} +
      > frac{delta^4}{3} - frac{delta^5}{4} + ... quad quad $       (b)



      Adding Eq. a and b and taking $ ln(1+delta) $ common, we get,



      $ (1+delta) ln(1+delta) = delta + frac{delta^2}{2} -
      > frac{delta^3}{6} + ... $



      Ignoring higher order terms,



      $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
      > frac{delta^3}{6} $



      We can further write it as,



      $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
      > frac{delta.delta^2}{6} $



      And take an assumption such that,



      $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
      > frac{delta^2}{6} $



      $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{3} $



      Now, from Eq. 2,



      $ {(1 + delta)^{1 + delta}} = e^{(1+delta)ln(1+delta)} =
      > e^{delta+ frac{delta^2}{3}} $



      Therefore, Eq. 1 becomes,



      $ P( X geq (1+delta)mu) leq [frac {e^{delta}} {(1 + delta)^{1 +
      > delta}}]^{mu}= [frac {e^{delta}} {e^{delta+
      > frac{delta^2}{3}}}]^{mu} = [ {e^{delta - delta-
      > frac{delta^2}{3}}}]^{mu} $



      $ P( X geq (1+delta)mu) leq {e^{-frac{delta^2}{3}}}^{mu} $







      probability-theory inequality







      share|cite|improve this question













      How can we prove
      $latex P( X geq (1+delta)mu) leq {e}^{frac{-delta}{3}{mu}} quad where quad delta > 1 $ -----------(1)



      When I have already proved that:
      $latex ( X geq (1+delta)mu) leq {e}^{frac{-delta^2}{3}{mu}} quad where quad 0 leq delta leq 1 $ -----------(2)



      How can i extend the second Prof to show that (1) is also true.




      Solution to 2



      $ P( X geq (1+delta)mu) leq [frac {e^{delta}} {(1 + delta)^{1 +
      > delta}}]^{mu} leq {e}^{frac{-delta^2}{3}{mu}} quad where quad
      > 0 leq delta leq 1 quad quad $       (1)



      Consider the denominator part to be,



      $ {(1 + delta)^{1 + delta}} = e^{(1+delta)ln(1+delta)} quad
      > quad $      (2)



      We can apply Taylor’s Series on,



      $ {(1+delta)ln(1+delta)} quad As quad delta=[0,1] $



      Therefore,



      $ ln(1+delta) = delta - frac{delta^2}{2} + frac{delta^3}{3} -
      > frac{delta^4}{4} + ... quad quad $      (a)



      $ delta ln(1+delta) = delta^2 - frac{delta^3}{2} +
      > frac{delta^4}{3} - frac{delta^5}{4} + ... quad quad $       (b)



      Adding Eq. a and b and taking $ ln(1+delta) $ common, we get,



      $ (1+delta) ln(1+delta) = delta + frac{delta^2}{2} -
      > frac{delta^3}{6} + ... $



      Ignoring higher order terms,



      $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
      > frac{delta^3}{6} $



      We can further write it as,



      $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
      > frac{delta.delta^2}{6} $



      And take an assumption such that,



      $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{2} -
      > frac{delta^2}{6} $



      $ (1+delta) ln(1+delta) geq delta + frac{delta^2}{3} $



      Now, from Eq. 2,



      $ {(1 + delta)^{1 + delta}} = e^{(1+delta)ln(1+delta)} =
      > e^{delta+ frac{delta^2}{3}} $



      Therefore, Eq. 1 becomes,



      $ P( X geq (1+delta)mu) leq [frac {e^{delta}} {(1 + delta)^{1 +
      > delta}}]^{mu}= [frac {e^{delta}} {e^{delta+
      > frac{delta^2}{3}}}]^{mu} = [ {e^{delta - delta-
      > frac{delta^2}{3}}}]^{mu} $



      $ P( X geq (1+delta)mu) leq {e^{-frac{delta^2}{3}}}^{mu} $







      probability-theory inequality




      probability-theory inequality






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 21 at 21:43









      Muhammad Fasiurrehman Sohi

      137




      137



























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