# absolute moments bounding (necessary and sufficient condition)

Let $X$ be a random variable  ; then

 $E[\left|X\right|^{k}]\leq M^{k}\text{ \ \ \ \ \ \ }\forall k\geq 1,k\in\mathbf% {N}$

if and only if,$\forall i\geq 0,i\in\mathbf{N}$

 $E\left[\left|X\right|^{k}\right]\leq E\left[\left|X\right|^{i}\right]M^{k-i}% \text{ \ \ \ \ \ \ \ }\forall k\geq i,k\in\mathbf{N}$

###### Proof.

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

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

b) $(E[\left|X\right|^{k}]\leq M^{k}$ $\Longrightarrow E\left[\left|X\right|^{k}\right]\leq E\left[\left|X\right|^{i}% \right]M^{k-i})$

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

 $\displaystyle E[\left|X\right|^{k}]$ $\displaystyle=$ $\displaystyle E\left[\left|X\right|^{\frac{i}{2}}\left|X\right|^{k-\frac{i}{2}% }\right]$ $\displaystyle\leq$ $\displaystyle E\left[\left|X\right|^{i}\right]^{\frac{1}{2}}E\left[\left|X% \right|^{2k-i}\right]^{\frac{1}{2}}$ $\displaystyle=$ $\displaystyle E\left[\left|X\right|^{i}\right]^{\frac{1}{2}}E\left[\left|X% \right|^{\frac{i}{2}}\left|X\right|^{2k-\frac{3}{2}i}\right]^{\frac{1}{2}}$ $\displaystyle\leq$ $\displaystyle E\left[\left|X\right|^{i}\right]^{\left(\frac{1}{2}+\frac{1}{4}% \right)}E\left[\left|X\right|^{4k-3i}\right]^{\frac{1}{4}}$ $\displaystyle\leq$ $\displaystyle E\left[\left|X\right|^{i}\right]^{\left(\frac{1}{2}+\frac{1}{4}+% \frac{1}{8}\right)}E\left[\left|X\right|^{\left(8k-7i\right)}\right]^{\frac{1}% {8}}$ $\displaystyle...$ $\displaystyle\leq$ $\displaystyle E\left[\left|X\right|^{i}\right]^{\left(\sum_{m=1}^{N}\frac{1}{2% ^{m}}\right)}E\left[\left|X\right|^{2^{N}k-\left(2^{N}-1\right)i}\right]^{% \frac{1}{2^{N}}}$ $\displaystyle=$ $\displaystyle E\left[\left|X\right|^{i}\right]^{\left(1-\frac{1}{2^{N}}\right)% }E\left[\left|X\right|^{2^{N}\left(k-i\right)+i}\right]^{\frac{1}{2^{N}}}$ $\displaystyle\leq$ $\displaystyle E\left[\left|X\right|^{i}\right]^{\left(1-\frac{1}{2^{N}}\right)% }M^{\left(k-i\right)+\frac{i}{2^{N}}},$

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

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

Title absolute moments bounding (necessary and sufficient condition) AbsoluteMomentsBoundingnecessaryAndSufficientCondition 2013-03-22 16:13:58 2013-03-22 16:13:58 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 5 Andrea Ambrosio (7332) Theorem msc 60E15