absolute moments bounding (necessary and sufficient condition)


Let X be a random variableMathworldPlanetmath; then

E[|X|k]Mk k1,k𝐍

if and only if,i0,i𝐍

E[|X|k]E[|X|i]Mk-i ki,k𝐍

Proof.

a) (E[|X|k]E[|X|i]Mk-i    E[|X|k]Mk)

It’s enough to take i=0 and the thesis follows easily.


b) (E[|X|k]Mk E[|X|k]E[|X|i]Mk-i)

Let 1ik (the case i=0 is trivial). Then, using Cauchy-Schwarz inequality N times, one has:

E[|X|k] = E[|X|i2|X|k-i2]
E[|X|i]12E[|X|2k-i]12
= E[|X|i]12E[|X|i2|X|2k-32i]12
E[|X|i](12+14)E[|X|4k-3i]14
E[|X|i](12+14+18)E[|X|(8k-7i)]18
E[|X|i](m=1N12m)E[|X|2Nk-(2N-1)i]12N
= E[|X|i](1-12N)E[|X|2N(k-i)+i]12N
E[|X|i](1-12N)M(k-i)+i2N,

and since this must hold for any N, we obtain

E[|X|k]E[|X|i]Mk-i

Title absolute moments bounding (necessary and sufficient condition)
Canonical name AbsoluteMomentsBoundingnecessaryAndSufficientCondition
Date of creation 2013-03-22 16:13:58
Last modified on 2013-03-22 16:13:58
Owner Andrea Ambrosio (7332)
Last modified by Andrea Ambrosio (7332)
Numerical id 5
Author Andrea Ambrosio (7332)
Entry type Theorem
Classification msc 60E15