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

1) $\Pr\left\{\left|X\right|\leq M\right\}=1$  i.e. $X$ is absolutely bounded almost surely;

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

Proof.

1) $\Longrightarrow$ 2)

Let’s define

 $F=\left\{\omega\in\Omega:\left|X\left(\omega\right)\right|>M\right\};$
 $\Pr\left\{\Omega\backslash F\right\}=1$

and

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

We have:

 $\displaystyle E[\left|X\right|^{k}]$ $\displaystyle=$ $\displaystyle\int_{\Omega}\left|X\right|^{k}dP$ $\displaystyle=$ $\displaystyle\int_{\Omega\backslash F}\left|X\right|^{k}dP+\int_{F}\left|X% \right|^{k}dP$ $\displaystyle=$ $\displaystyle\int_{\Omega\backslash F}\left|X\right|^{k}dP$ $\displaystyle\leq$ $\displaystyle\int_{\Omega\backslash F}M^{k}dP$ $\displaystyle=$ $\displaystyle M^{k}\Pr\left\{\Omega\backslash F\right\}=M^{k}.$

2) $\Longrightarrow$ 1)

Let’s define

 $\displaystyle F$ $\displaystyle=$ $\displaystyle\left\{\omega\in\Omega:\left|X\left(\omega\right)\right|>M\right\}$ $\displaystyle F_{n}$ $\displaystyle=$ $\displaystyle\left\{\omega\in\Omega:\left|X\left(\omega\right)\right|>M+\frac{% 1}{n}\right\}\text{ \ }\forall n\geq 1.$

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

 $F=\lim_{n\rightarrow\infty}F_{n}$

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

 $\Pr\left\{F\right\}=\Pr\left\{\lim_{n\rightarrow\infty}F_{n}\right\}=\lim_{n% \rightarrow\infty}\Pr\left\{F_{n}\right\}.$

Now, for any $k\geq 1$,

 $\displaystyle M^{k}$ $\displaystyle\geq$ $\displaystyle E\left[\left|X\right|^{k}\right]$ $\displaystyle=$ $\displaystyle\int_{\Omega}\left|X\left(\omega\right)\right|^{k}dP$ $\displaystyle=$ $\displaystyle\int_{\Omega\backslash F_{n}}\left|X\left(\omega\right)\right|^{k% }dP+\int_{F_{n}}\left|X\left(\omega\right)\right|^{k}dP$ $\displaystyle\geq$ $\displaystyle\int_{F_{n}}\left|X\left(\omega\right)\right|^{k}dP$ $\displaystyle\geq$ $\displaystyle\int_{F_{n}}\left(M+\frac{1}{n}\right)^{k}dP$ $\displaystyle=$ $\displaystyle\left(M+\frac{1}{n}\right)^{k}\Pr\left\{F_{n}\right\}.$

that is

 $\Pr\left\{F_{n}\right\}\leq\left(\frac{M}{M+\frac{1}{n}}\right)^{k}\text{ \ for any }k\geq 1$

so that the only acceptable value for $\Pr\left\{F_{n}\right\}$ is

 $\Pr\left\{F_{n}\right\}=0$

whence the thesis. ∎

Acknowledgements: due to helpful discussions with Mathprof.

Title relation between almost surely absolutely bounded random variables and their absolute moments RelationBetweenAlmostSurelyAbsolutelyBoundedRandomVariablesAndTheirAbsoluteMoments 2013-03-22 16:14:33 2013-03-22 16:14:33 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 8 Andrea Ambrosio (7332) Theorem msc 60A10