absolute moments bounding (necessary and sufficient condition)


Let X be a random variableMathworldPlanetmath; then

E⁢[|X|k]≤Mk⁢ ⁢∀k≥1,k∈𝐍

if and only if,∀i≥0,i∈𝐍

E⁢[|X|k]≤E⁢[|X|i]⁢Mk-i⁢ ⁢∀k≥i,k∈𝐍

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 1≤i≤k (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]12⁢E⁢[|X|2⁢k-i]12
= E⁢[|X|i]12⁢E⁢[|X|i2⁢|X|2⁢k-32⁢i]12
≤ E⁢[|X|i](12+14)⁢E⁢[|X|4⁢k-3⁢i]14
≤ E⁢[|X|i](12+14+18)⁢E⁢[|X|(8⁢k-7⁢i)]18
…
≤ E⁢[|X|i](∑m=1N12m)⁢E⁢[|X|2N⁢k-(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