absolute moments bounding (necessary and sufficient condition)

Let $X$ be a random variable; then

$$E[{\left|X\right|}^k]\le M^k\forall k\ge 1,k\in 𝐍$$

if and only if,$\forall i\ge 0,i\in 𝐍$

$$E\left[{\left|X\right|}^k\right]\le E\left[{\left|X\right|}^i\right]M^{k-i}\forall k\ge i,k\in 𝐍$$

Proof.

a) $(E\left[\right|X{|}^k]\le E\left[\right|X{|}^i]M^{k-i}$  $⟹$  $E[|X{|}^k]\le M^k)$

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

b) $(E[|X{|}^k]\le M^k$ $⟹E\left[\right|X{|}^k]\le E\left[\right|X{|}^i]M^{k-i})$

Let $1\le i\le k$ (the case $i=0$ is trivial). Then, using Cauchy-Schwarzinequality $N$ times, one has:

$E[{\left|X\right|}^k]$$=$$E\left[{\left|X\right|}^{\frac{i}{2}}{\left|X\right|}^{k-\frac{i}{2}}\right]$

$\le$$E{\left[{\left|X\right|}^i\right]}^{\frac{1}{2}}E{\left[{\left|X\right|}^{2k-i}\right]}^{\frac{1}{2}}$

$=$$E{\left[{\left|X\right|}^i\right]}^{\frac{1}{2}}E{\left[{\left|X\right|}^{\frac{i}{2}}{\left|X\right|}^{2k-\frac{3}{2}i}\right]}^{\frac{1}{2}}$

$\le$$E{\left[{\left|X\right|}^i\right]}^{\left(\frac{1}{2}+\frac{1}{4}\right)}E{\left[{\left|X\right|}^{4k-3i}\right]}^{\frac{1}{4}}$

$\le$$E{\left[{\left|X\right|}^i\right]}^{\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)}E{\left[{\left|X\right|}^{\left(8k-7i\right)}\right]}^{\frac{1}{8}}$

$\mathrm{\dots }$

$\le$$E{\left[{\left|X\right|}^i\right]}^{\left({\sum }_{m=1}^N\frac{1}{2^m}\right)}E{\left[{\left|X\right|}^{2^{N}k-\left(2^N-1\right)i}\right]}^{\frac{1}{2^N}}$

$=$$E{\left[{\left|X\right|}^i\right]}^{\left(1-\frac{1}{2^N}\right)}E{\left[{\left|X\right|}^{2^N\left(k-i\right)+i}\right]}^{\frac{1}{2^N}}$

$\le$$E{\left[{\left|X\right|}^i\right]}^{\left(1-\frac{1}{2^N}\right)}{M}^{\left(k-i\right)+\frac{i}{2^N}},$

and since this must hold for any $N$, we obtain

$$E[{\left|X\right|}^k]\le E\left[{\left|X\right|}^i\right]M^{k-i}$$

∎