AbsoluteMomentsBoundingNecessaryAndSufficientCondition 2006-09-09 10:31:18 2006-09-09 10:32:16 Theorem % 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 $X$ be a random variable; then% \[ E[\left\vert X\right\vert ^{k}]\leq M^{k}\text{ \ \ \ \ \ \ }\forall k\geq 1,k\in \mathbf{N} \] \bigskip if and only if,$\forall i\geq 0,i\in \mathbf{N}$ \[ E\left[ \left\vert X\right\vert ^{k}\right] \leq E\left[ \left\vert X\right\vert ^{i}\right] M^{k-i}\text{ \ \ \ \ \ \ \ }\forall k\geq i,k\in \mathbf{N} \] \bigskip \begin{proof} a) $(E\left[ \left\vert X\right\vert ^{k}\right] \leq E\left[ \left\vert X\right\vert ^{i}\right] M^{k-i}$ \ $\Longrightarrow $ \ $% E[\left\vert X\right\vert ^{k}]\leq M^{k})$ It's enough to take $i=0$ and the thesis follows easily. \bigskip b) $(E[\left\vert X\right\vert ^{k}]\leq M^{k}$ $\Longrightarrow E\left[ \left\vert X\right\vert ^{k}\right] \leq E\left[ \left\vert X\right\vert ^{i}% \right] M^{k-i})$ Let $1\leq i\leq k$ (the case $i=0$ is trivial). Then, using Cauchy-Schwarz inequality $N$ times, one has:% \begin{eqnarray*} E[\left\vert X\right\vert ^{k}] &=&E\left[ \left\vert X\right\vert ^{\frac{i% }{2}}\left\vert X\right\vert ^{k-\frac{i}{2}}\right] \\ &\leq &E\left[ \left\vert X\right\vert ^{i}\right] ^{\frac{1}{2}}E\left[ \left\vert X\right\vert ^{2k-i}\right] ^{\frac{1}{2}} \\ &=&E\left[ \left\vert X\right\vert ^{i}\right] ^{\frac{1}{2}}E\left[ \left\vert X\right\vert ^{\frac{i}{2}}\left\vert X\right\vert ^{2k-\frac{3}{2% }i}\right] ^{\frac{1}{2}} \\ &\leq &E\left[ \left\vert X\right\vert ^{i}\right] ^{\left( \frac{1}{2}+% \frac{1}{4}\right) }E\left[ \left\vert X\right\vert ^{4k-3i}\right] ^{\frac{1% }{4}} \\ &\leq &E\left[ \left\vert X\right\vert ^{i}\right] ^{\left( \frac{1}{2}+% \frac{1}{4}+\frac{1}{8}\right) }E\left[ \left\vert X\right\vert ^{\left( 8k-7i\right) }\right] ^{\frac{1}{8}} \\ &&... \\ &\leq &E\left[ \left\vert X\right\vert ^{i}\right] ^{\left( \sum_{m=1}^{N}% \frac{1}{2^{m}}\right) }E\left[ \left\vert X\right\vert ^{2^{N}k-\left( 2^{N}-1\right) i}\right] ^{\frac{1}{2^{N}}} \\ &=&E\left[ \left\vert X\right\vert ^{i}\right] ^{\left( 1-\frac{1}{2^{N}}% \right) }E\left[ \left\vert X\right\vert ^{2^{N}\left( k-i\right) +i}\right] ^{\frac{1}{2^{N}}} \\ &\leq &E\left[ \left\vert X\right\vert ^{i}\right] ^{\left( 1-\frac{1}{2^{N}}% \right) }M^{\left( k-i\right) +\frac{i}{2^{N}}}, \end{eqnarray*} and since this must hold for any $N$, we obtain \[ E[\left\vert X\right\vert ^{k}]\leq E\left[ \left\vert X\right\vert ^{i}% \right] M^{k-i} \] \end{proof}