Conditional Expectations of Random Variables Without Expectations

Abstract

Let $(\Omega, \mathscr{F}, P)$ be a probability space, and let $X$ be a random variable defined on $(\Omega, \mathscr{F}, P)$. If $\mathscr{A}$ is a sub $\sigma$-field of $\mathscr{F}$, then $E(X \mid \mathscr{A})$ is the a.s. unique $\mathscr{A}$ measurable function such that, for all $A \varepsilon \mathscr{A}$, \begin{equation*}\tag{1}\int_A X dP = \int_A E(X \mid \mathscr{A}) dP,\end{equation*} provided $EX$ is defined. ([2], p. 341). If $EX$ is not defined, that is, if $EX^+ = EX^- = \infty$, we may then define $E(X \mid \mathscr{A}) = E(X^+ \mid \mathscr{A}) - E(X^- \mid \mathscr{A})$, provided the difference is defined almost surely ([2], p. 342). We show that this is the only reasonable definition of $E(X \mid \mathscr{A})$ (Lemma 2), and exhibit several apparent pathologies, akin to the fact that a conditionally convergent series of real numbers may be re-ordered to give any sum. If $X$ is any random variable with a continuous distribution such that $EX$ is not defined, then we can find an $\mathscr{A} \subset \mathscr{F}$ such that $E(X \mid \mathscr{A}) = 0$ a.s. (Theorem 1), and if $Y$ is any random variable independent of $X$, we can find an $\mathscr{A} \subset \mathscr{F}$ such that $E(X \mid \mathscr{A}) = Y$ a.s. (Theorem 2). In fact, if $X_1, X_2, \cdots$ is a sequence of independent random variables such that for $n \geqq 2, EX_n$ is not defined and $X_n$ has a continuous distribution, we can find a sequence of $\sigma$-fields $\mathscr{A}_1 \subset \mathscr{A}_2 \subset \cdots \subset \mathscr{F}$ such that $X_1, \cdots, X_n$ are $\mathscr{A}_n$ measurable and $E(X_{n + 1} \mid \mathscr{A}_n) = X_n$ a.s. (Theorem 3). The sequence $\{X_n, \mathscr{F}_n, n = 1, 2, \cdots\}$ is not a martingale however, since for $m > n + 1, E(X_m \mid \mathscr{F}_n)$ is not defined. We remark that the standard theorem on iterated conditional expectations, ([2], p. 350) which says that if $\mathscr{A} \subset \mathscr{B} \subset \mathscr{F}$ then $E(X \mid \mathscr{A}) = E(E(X \mid \mathscr{B}) \mid \mathscr{A})$ a.s. is valid only if $E(X \mid \mathscr{A})$ is defined a.s.