Mixing
Chernoff Bound Variantion.
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
asked Nov 21 '18 at 21:43
Muhammad Fasiurrehman SohiMuhammad Fasiurrehman Sohi
187
add a comment |
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
asked Nov 21 '18 at 21:43
Muhammad Fasiurrehman SohiMuhammad Fasiurrehman Sohi
187
add a comment |
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
asked Nov 21 '18 at 21:43
Muhammad Fasiurrehman SohiMuhammad Fasiurrehman Sohi
187
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
asked Nov 21 '18 at 21:43
Muhammad Fasiurrehman SohiMuhammad Fasiurrehman Sohi
187
asked Nov 21 '18 at 21:43
Muhammad Fasiurrehman SohiMuhammad Fasiurrehman Sohi
187
asked Nov 21 '18 at 21:43
Muhammad Fasiurrehman SohiMuhammad Fasiurrehman Sohi
187
asked Nov 21 '18 at 21:43
Muhammad Fasiurrehman SohiMuhammad Fasiurrehman Sohi
187
asked Nov 21 '18 at 21:43
Muhammad Fasiurrehman SohiMuhammad Fasiurrehman Sohi
187
187
add a comment |
add a comment |
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