RelationBetweenAlmostSurelyAbsolutelyBoundedRandomVariablesAndTheirAbsoluteMoments 2006-09-13 16:25:35 2006-09-16 13:29:04 Theorem almost surely absolutely bounded random variable % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $\{\Omega ,E,P\}$ a probability space and let $X$ be a random variable; then, the following are equivalent: 1) $\Pr \left\{ \left\vert X\right\vert \leq M\right\} =1$ \ i.e. $X$ is absolutely bounded almost surely; 2) $E[\left\vert X\right\vert ^{k}]\leq M^{k}$ \ \ \ \ \ \ $\forall k\geq 1,k\in N$ \begin{proof} 1) $\Longrightarrow $ 2) Let's define% \[ F=\left\{ \omega \in \Omega :\left\vert X\left( \omega \right) \right\vert >M\right\} ; \] Then by hypothesis \[ \Pr \left\{ \Omega \backslash F\right\} =1 \] and \[ \Pr \left\{ F\right\} =0. \] We have: \begin{eqnarray*} E[\left\vert X\right\vert ^{k}] &=&\int_{\Omega }\left\vert X\right\vert ^{k}dP \\ &=&\int_{\Omega \backslash F}\left\vert X\right\vert ^{k}dP+\int_{F}\left\vert X\right\vert ^{k}dP \\ &=&\int_{\Omega \backslash F}\left\vert X\right\vert ^{k}dP \\ &\leq &\int_{\Omega \backslash F}M^{k}dP \\ &=&M^{k}\Pr \left\{ \Omega \backslash F\right\} =M^{k}. \end{eqnarray*} 2) $\Longrightarrow $ 1) Let's define \begin{eqnarray*} F &=&\left\{ \omega \in \Omega :\left\vert X\left( \omega \right) \right\vert >M\right\} \\ F_{n} &=&\left\{ \omega \in \Omega :\left\vert X\left( \omega \right) \right\vert >M+\frac{1}{n}\right\} \text{ \ }\forall n\geq 1. \end{eqnarray*} Then we have obviously $F_{n}\subseteq F_{n+1}$ (in fact, if $\omega \in F_{n}\Longrightarrow \left\vert X\left( \omega \right) \right\vert >M+\frac{1% }{n}>M+\frac{1}{n+1}\Longrightarrow \omega \in F_{n+1}$) and $% F=\bigcup_{n=1}^{\infty }F_{n}$ (in fact, let $\omega \in F$; let $% N=\left\lceil \frac{1}{\left\vert X\left( \omega \right) \right\vert -M}% \right\rceil $; then $\left\vert X\left( \omega \right) \right\vert >M+\frac{% 1}{N}$, that is $\omega \in F_{N}$); this means that% \[ F=\lim_{n\rightarrow \infty }F_{n} \] in the meaning of \PMlinkname{sets sequences convergence}{SequenceOfSetsConvergence}. So \PMlinkname{the continuity from below property}{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 $k\geq 1$, \begin{eqnarray*} M^{k} &\geq &E\left[ \left\vert X\right\vert ^{k}\right] \\ &=&\int_{\Omega }\left\vert X\left( \omega \right) \right\vert ^{k}dP \\ &=&\int_{\Omega \backslash F_{n}}\left\vert X\left( \omega \right) \right\vert ^{k}dP+\int_{F_{n}}\left\vert X\left( \omega \right) \right\vert ^{k}dP \\ &\geq &\int_{F_{n}}\left\vert X\left( \omega \right) \right\vert ^{k}dP \\ &\geq &\int_{F_{n}}\left( M+\frac{1}{n}\right) ^{k}dP \\ &=&\left( M+\frac{1}{n}\right) ^{k}\Pr \left\{ F_{n}\right\} . \end{eqnarray*} 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. \end{proof} Acknowledgements: due to helpful discussions with Mathprof.