% spellchecked: 10-11-98pdh % last edit: 10-11-98pdh % gallies sent: % gallies corrected: % set in production style: 10-11-98pdh % Section Inroads editor Bullen % Received 4/20/98 % AMS # 26A39, 60G46 \documentclass{rae} \usepackage{amsmath,amsthm,amssymb,enumerate} %\coverauthor{Valentin Skvortsov} %\covertitle{A Martingale Closure Theorem for $A$-Integrable Martingale Sequences} \received{April 20, 1998} \MathReviews{26A39, 60G46} \keywords {A-integral, Uniform A-integrability, Martingale sequence} \firstpagenumber{1} \markboth{Valentin~Skvortsov}{A Martingale Closure Theorem} \author{Valentin Skvortsov, Department of Mathematics, Moscow State University, Moscow, 119899, Russia, e-mail: {\tt vaskvor@nw.math.msu.su} or: Institute of Mathematics, WSP, Plac Weyssenhoffa 11, Bydgoszcz, 85-072, Poland, e-mail: {\tt skworcow@wsp.bydgoszcz.pl}} \title{MARTINGALE CLOSURE THEOREM FOR $A$-INTEGRABLE MARTINGALE SEQUENCES} %%%%%%Put Author's Definitions Below Here %%%%%%%%%%% \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \theoremstyle{definition} \newtheorem{definition}{Definition} \newtheorem{remark}{Remark} \def\F{{\cal F}} \def\X{{X_\infty}} \def\ep{{\varepsilon}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \maketitle \begin{abstract} A generalized conditional expectation and the corresponding martingale is defined in terms of the Kolmogorov $A$-integral. It is proved that the uniform $A$-integrability of a martingale sequence is a sufficient condition for the sequence to be closed on the right by the $A$-integrable last element. \end{abstract} \bigskip A well known theorem in martingale theory states that a martingale sequence $\{X_n,\F _n$, $n=1,2,\ldots\}$ is closed on the right by the last element $\X (\omega)=\lim\limits_{n\to\infty} X_n(\omega)$ iff >$\{X_n,\F _n\}$ is a uniformly integrable sequence (see \cite[p.300]{ash}, \cite[p.239]{cho} or \cite[p.60]{loe}). The conditional expectation in this theory is defined in terms of the Lebesgue integral. Meanwhile there are some other versions of a notion of the mathematical expectation which involve integration more general than the Lebesgue integration. One of such generalization was introduced by Kolmogorov in \cite{kol} who defined generalized mathematical expectation as a non-absolutely convergent integral which later became known as the Kolmogorov $A$-integral (see \cite{bar}, \cite{lee}). In this note we are extending this definition to the case of the conditional expectation and applying this extension to the investigation of $A$-integrable martingales. We recall some definitions. \definition{ A random variable (r.~v.) $X$ defined on a probability space $(\Omega, {\cal{B}}, P)$ is said to be {\it $A$-integrable} over a set $B\in {\cal{B}}$ if $$P\bigl\{\omega\in B\colon |X(\omega)|>C\bigr\}=\bar{o}(1/C) \mbox{ as } C\to\infty$$ and if there exists a finite limit $$\lim\limits_{C\to\infty}\!\!\!\! \int\limits_{\{\omega\in B\colon\: |X(\omega)|\le C\}} \!\!\!\!\!\!\!\!\!\!\!\!X\, dP \,= \,I.$$ Then $I$ is called {\it the $A$-integral} of $X$ over $B$ and is denoted by $(A)\int\limits_B X\,dP$. Note that if a r.~v. $X$ is $A$-integrable over some set $B\in {\cal{B}}$ and $X(\omega)\ge 0$ a.~s. on $B$ then $X$ is also $L$-integrable and $(A)\int\limits_B X\,dP= (L)\int\limits_B X\,dP$. This fact implies that if an $\F$-measurable r.~v. $X$ ($\F$ being a sub-$\sigma$-field of ${\cal{B}}$) is $A$-integrable over any $\F$-measurable subset of some $\F$-measurable set $B$ then $X$ is $L$-integrable over $B$. Indeed, put $B_{+}=\{\omega\in B\colon X(\omega)\ge 0\}$ and $B_{-}=\{\omega\in B\colon X(\omega)<0\}$. Then, being the r.~v. $X$ $A$-integrable over $B_{+}$ and over $B_{-}$, it must be $L$-integrable over each of these sets and consequently over $B$. We use this observation in the following definition. \definition{ Let a r.~v. $X$ be $A$-integrable over any set $B\in\F$ where $\F$ is a sub-$\sigma$-field of ${\cal B}$. {\it The conditional $A$-expectation of $X$ with respect to $\F$} is defined as an $\F$-measurable r.~v. $AE(X|F)$ such that for every $B\in\F$ we have $$\label{exp} \int\limits_B AE(X|\F)\, dP=(A)\int\limits_B X\, dP.$$ } \definition{ An $A$-integrable r.~v. $\X$ is said to be {\it the last element} of a martingale sequence $\{X_n,\F_n, n=1,2,\ldots\}$ if $X_n=AE(\X|\F_n)$ for each $n=1,2,\ldots$. We also say in this case that $\X$ {\it closes the martingale sequence from the right}. } We have omitted the "$A$"-sign in front of the left hand side of (\ref{exp}) because of the above observation, meaning that the Lebesgue integral can be used here. It follows from the same observation that the use of the $A$-integral in the definition of a martingale sequence gives an essentially more general notion only for the last element and only in the case where each $\sigma$-field $\F _n$ is a proper subset of the $\sigma$-field $\F _\infty$generated by $\cup_n\F_n$. Here we are going to give a sufficient condition for a martingale sequence to be closed on the right by the $A$-integrable last element. This condition is formulated in terms of uniform $A$-integrability which is a non-absolute analogue of the uniform Lebesgue integrability. \definition{ A family of r. v. $\{X_\gamma\}_{\gamma\in\Gamma}$, defined on $(\Omega, {\cal{B}}, P)$ ($\Gamma$ is some index set) is said to be {\it uniformly $A$-integrable on $B\in {\cal{B}}$} iff $$\sup\limits_{\gamma\in\Gamma}\; \Biggl|\: (A)\int\limits_{D_{\gamma}(C)} X_\gamma\, dP\:\Biggr|\longrightarrow 0\quad \mbox{ as } C\to\infty,$$ where $D_{\gamma}(C)=\{\omega\in B\colon\,|X_\gamma(\omega)|>C\}$. } Now let $\{\F _n\}$ be an increasing sequence of sub-$\sigma$-fields of ${\cal B}$ and let $\{X_n, \F _n$, $n=1,2,\ldots\}$ be a martingale. For any $B\in\cup _n\F _n$ we define a set function $\Phi$ by putting $$\label{phi} \Phi(B)=\int\limits_B X_n\, dP \quad \mbox{ if } B\in\F _n.$$ We call $\Phi$ {\it the associated set function} for $\{X_n,\F _n\}$. Note that $\Phi$ is well defined. Indeed, if $m\ge n$ then for the same $B$ by (\ref{phi}) with $n$ substituted by $m$ we get $$\label{phim} \Phi(B)=\int\limits_B X_m\, dP \quad \mbox{ if } B\in\F _n\subseteq\F _m.$$ Now by the definition of the martingale and by the definition of the conditional expectation we get $$\int\limits_B X_m\, dP=\int\limits_B E(X_m|\F _n)\, dP=\int\limits_B X_n\, dP$$ and this proves that the values of $\Phi$ on $B$ given by (\ref{phi}) and (\ref{phim}) coincide. $\Phi$ is of course additive on $\cup _n\F _n$ but we do not assume that $\Phi$ can be extended to the $\sigma$-field $\F _\infty$ generated by $\cup _n\F _n$. \begin{lemma} A r.~v. $\X$ is the last element of a martingale $\{X_n, \F _n, n=1,2,\ldots,\infty\}$ in the sense of the $A$-integral iff for the associated set function $\Phi$ of the martingale sequence $\{X_n,\F _n, n=1,2,\ldots\}$ we have $(A)\int _B \X\, dP=\Phi(B)$ for any $B\in\cup _n\F _n$. \end{lemma} \begin{proof} This follows directly from the definition of $\Phi$ and from the definition of the last element.\end{proof} Note that this Lemma is true for any other integral which can be used in the above equality. \begin{theorem} Let $\{X_n,\F _n, n=1,2,\ldots\}$ be a martingale sequence convergent a.~s. to a r.~v. $\X$. Let $$P\bigl\{\omega\in\Omega\colon |X_n(\omega)|>C\bigr\}=\bar{o}(1/C) \mbox{ uniformly in n as } C\to\infty$$ and $\{X_n\}$ be uniformly $A$-integrable on any $B\in\cup _n\F _n$ in the sense of Definition 3. Then $\X$ is $A$-integrable on each $B\in\cup _n\F_n$ and closes the martingale sequence $\{X_n,\F _n\}$ on the right, i.~e. $\{X_n,\F _n, n=1,2,\ldots,\infty\}$ is an $A$-integrable martingale with $$X_n=AE(\X|\F _n)$$ $\X$ being its last element. \end{theorem} \begin{proof} We show first that (5) implies (1) with $X=\X$. It is enough to prove (1) with $B=\Omega$. \ Denote \ $D(C)=\bigl\{\omega\in\Omega\colon |\X (\omega)|>C\bigr\},\quad D_n(C)=$ $\bigl\{\omega\in\Omega\colon |X_n(\omega)|>C\bigr\},\quad G_n(C)=\bigcap\limits_{m=n}^\infty D_m(C)$. Then obviously\ $P(D(C))\le P(\bigcup\limits_{n=1}^\infty G_n(C)),\quad G_n(C)\subseteq G_{n+1}(C),$ $$P\bigl(D(C)\bigr)\le \lim\limits_{n\to\infty} P\bigl(G_n(C)\bigr),$$ $$G_n(C)\subseteq D_n(C).$$ Fix any $\ep >0$. By (5) there exists $C_{\ep}>0$ such that $P\bigl(D_n(C)\bigr)\le \ep/C$ for all $n=1,2\ldots$ and for any $C\ge C_{\ep}$. Fix such $C$. Then by (6) and (7) $P\bigl(D(C)\bigr)\le \ep/C$. As $\ep >0$ is arbitrary and $C$ is such that $C\ge C_{\ep}$, then (1) with $X=\X$ is proved for $B=\Omega$ and therefore for any $B\in {\cal{B}}$. For any r. v. $X$ and $C>0$ define $$X^C(\omega)= \left\{ \begin{array}{rcl} X(\omega),& & \mbox{ if }\: |X(\omega)|\le C,\\ C \operatorname{sign} X(\omega),& & \mbox{otherwise}.\\ \end{array} \right.$$ Notice that $$\lim\limits_{n\to\infty} X^C_n(\omega)=X_\infty^C(\omega)\quad \mbox{ a.~s. on } \Omega.$$ Now fix $B\in\cup _n\F _n$. Then $B\in \F _n$ for some $n$ and hence for the associated function $\Phi$ the equality (4) holds for any $m\ge n$. Fix $\ep >0$. Since the sequence $\{X_n,{\cal F}_n, n=1,2,\ldots\}$ is uniformly $A$-integrable on B and (5) holds, we can find $C_0$ such that for all $C\ge C_0$ and for all $m$ $$\Biggl|\: \int\limits_B \bigl(X_m-X_m^C\bigr)\, dP \:\Biggr|= \Biggl|\; \int\limits_{\{\omega\in B\colon\,|X_m(\omega)|>C\}}\!\!\!\!\!\!\!\!\!\!\!\!\!\! {X_m\, dP}\,$$ $$- C P{\bigl\{\omega\in B\colon\, X_m(\omega)>C\bigr\}}\, + C P{\bigl\{\omega\in B\colon\, X_m(\omega)<-C\bigr\}} \:\Biggr|\le$$ $$\le \Biggl|\:\int\limits_{\{\omega\in B\colon\,|X_m(\omega)|>C\}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! X_m \,dP\:\Biggr|+C P\bigl\{|X_m|>C\bigr\} < \ep/2$$ Let $C$ be also fixed for the moment. Then (8) and the Lebesgue dominated convergence theorem imply that for some $m=m_{\ep,C}\ge n$ $$\int\limits_B \bigl|X_m^C-X_\infty^C\bigr| \,dP < \ep/2$$ Now combining (4), (9) and (10) we get for the chosen $m$ $$\Biggl|\:\Phi(B)-\int\limits_B X_\infty^C \,dP\:\Biggr|= \Biggl|\: \int\limits_B X_m \, dP -\int\limits_B X_\infty^C \,dP \:\Biggr|\le$$ $$\le \Biggl|\: \int\limits_B \bigl(X_m-X_m^C\bigr) \,dP \:\Biggr| + \int\limits_B \bigl|X_m^C-X_\infty^C\bigr| \,dP < \ep$$ This together with (1) proved already for $X=\X$ implies that $\X$ is $A$-integrable on $B$ to $\Phi(B)$ and by Lemma, the r.~v. $\X$ is the last element, in the sense of the $A$-integral, of the considered martingale sequence. This completes the proof. \end{proof} Note that unlike in the case of the uniform Lebesgue integrability, the above condition in terms of the uniform $A$-integrability is not necessary for existence of the last element. This can be shown by constructing an example of a Haar series such that it is the $A$-Fourier series of an $A$-integrable function and its partial sums are not uniformly $A$-integrable. (See \cite{kos} for details.) \begin{thebibliography}{99} \bibitem{ash} {\sc R.~B.~Ash.}\quad Real Analysis and Probability,\quad Academic Press, New York, 1972. \bibitem{bar} {\sc N.~K.~Bary.}\quad A Treatise on Trigonometric Series,\quad Macmillan, New York, 1964. \bibitem{cho} {\sc Y.~S.~Chow, H.~Teicher.}\quad Probability Theory,\quad 2-nd edition, Springer-Verlag, 1988. \bibitem{kol} {\sc A.~N.~Kolmogorov.}\quad Foundations of the Theory of Probability,\quad Chelsea Publishing Company, New York, 1950. \bibitem{lee} {\sc L.~P.~Lee.}\quad Lanzhou Lectures on Henstock Integration,\quad World Scientific, Singapore, New Jersey, London, 1989. \bibitem{loe} {\sc M.~Loeve.}\quad Probability Theory II, \quad 4th edition, Springer-Verlag, New York, 1978. \bibitem{kos} {V.~A.~Skvortsov, V.~V.~Kostin.}\quad Martingale Sequences in the Theory of Orthogonal Series,\quad Vestnik of Moscow University, \quad in press. \end{thebibliography} \end{document}