relation between almost surely absolutely bounded random variables and their absolute moments

Let $\{\mathrm{\Omega },E,P\}$ a probability space and let $X$ be a randomvariable; then, the following are equivalent:

1) $\mathrm{Pr}\left\{\left|X\right|\le M\right\}=1$  i.e. $X$ isabsolutely bounded almost surely;

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

Proof.

1) $⟹$ 2)

Let’s define

$$F=\{\omega \in \mathrm{\Omega }:\left|X\left(\omega \right)\right|>M\};$$

Then by hypothesis

$$\mathrm{Pr}\left\{\mathrm{\Omega }\backslash \backslash F\right\}=1$$

and

$$\mathrm{Pr}\left\{F\right\}=0.$$

We have:

$E[{\left|X\right|}^k]$$=$${\displaystyle {\int }_{\mathrm{\Omega }}}{\left|X\right|}^k𝑑P$

$=$${\displaystyle {\int }_{\mathrm{\Omega }\backslash \backslash F}}{\left|X\right|}^k𝑑P+{\displaystyle {\int }_F}{\left|X\right|}^k𝑑P$

$=$${\displaystyle {\int }_{\mathrm{\Omega }\backslash \backslash F}}{\left|X\right|}^k𝑑P$

$\le$${\displaystyle {\int }_{\mathrm{\Omega }\backslash \backslash F}}{M}^k𝑑P$

$=$$M^k\mathrm{Pr}\left\{\mathrm{\Omega }\backslash \backslash F\right\}=M^k.$

2) $⟹$ 1)

Let’s define

$F$$=$$\{\omega \in \mathrm{\Omega }:\left|X\left(\omega \right)\right|>M\}$

$F_n$$=$$\{\omega \in \mathrm{\Omega }:\left|X\left(\omega \right)\right|>M+{\displaystyle \frac{1}{n}}\}\forall n\ge 1.$

Then we have obviously $F_n\subseteq F_{n+1}$ (in fact, if $\omega \in F_n⟹\left|X\left(\omega \right)\right|>M+\frac{1}{n}>M+\frac{1}{n+1}⟹\omega \in F_{n+1}$) and $F={\bigcup }_{n=1}^{\mathrm{\infty }}{F}_n$ (in fact, let $\omega \in F$; let $N=\lceil \frac{1}{\left|X\left(\omega \right)\right|-M}\rceil$; then $\left|X\left(\omega \right)\right|>M+\frac{1}{N}$, that is $\omega \in F_N$); this means that

$$F=\underset{n\to \mathrm{\infty }}{lim}{F}_n$$

in the meaning of sets sequences convergence (http://planetmath.org/SequenceOfSetsConvergence).

So the continuity from below property (http://planetmath.org/PropertiesForMeasure) of probability can be applied:

$$\mathrm{Pr}\left\{F\right\}=\mathrm{Pr}\left\{\underset{n\to \mathrm{\infty }}{lim}{F}_n\right\}=\underset{n\to \mathrm{\infty }}{lim}\mathrm{Pr}\left\{F_n\right\}.$$

Now, for any $k\ge 1$,

$M^k$$\ge$$E\left[{\left|X\right|}^k\right]$

$=$${\displaystyle {\int }_{\mathrm{\Omega }}}{\left|X\left(\omega \right)\right|}^k𝑑P$

$=$${\displaystyle {\int }_{\mathrm{\Omega }\backslash \backslash F_n}}{\left|X\left(\omega \right)\right|}^k𝑑P+{\displaystyle {\int }_{F_n}}{\left|X\left(\omega \right)\right|}^k𝑑P$

$\ge$${\displaystyle {\int }_{F_n}}{\left|X\left(\omega \right)\right|}^k𝑑P$

$\ge$${\displaystyle {\int }_{F_n}}{\left(M+{\displaystyle \frac{1}{n}}\right)}^k𝑑P$

$=$${\left(M+{\displaystyle \frac{1}{n}}\right)}^k\mathrm{Pr}\left\{F_n\right\}.$

that is

$$\mathrm{Pr}\left\{F_n\right\}\le {\left(\frac{M}{M+\frac{1}{n}}\right)}^k\text{ for any }k\ge 1$$

so that the only acceptable value for $\mathrm{Pr}\left\{F_n\right\}$ is

$$\mathrm{Pr}\left\{F_n\right\}=0$$

whence the thesis.∎

Acknowledgements: due to helpful discussions with Mathprof.