Htykuut

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} $