I assume that the reader acquainted with probability theory is familiar with conditional expectation, but for everyone’s benefit, we will briefly review the definition.
Definition. Let be a probability space, a sub--algebra, and be a random variable. A random varible such that
is called the conditional expectation of with respect to and denoted by .
If is the sub--algebra generated by a random variable , we often write instead of .
The definition of conditional expectation implicitly assumes that the random variable is a.s. unique, a result we prove now. Suppose satisfy the definition of conditional expectation. Then the sets and are in . Hence,
which implies that a.s. or equivalently, a.s.
Conditional expectation would not be such a useful concept unless it existed for a sufficiently large class of random variables. Fortunately, absolute integrability ensures the existence of conditional expectation of a random variable. The existence proof I first learned as a student used the Lebesgue-Radon-Nikodym theorem. There’s nothing wrong with this proof; any student of probability theory needs to understand absolute continuity and the Radon-Nikodym derivative. But the Radon-Nikodym route downplays the fact that conditional expectation is an operator on such that when , is the orthogonal projection of onto the subspace . Since I have a fondness for functional analysis–Hilbert spaces, in particular–I’m going to present an alternative proof from Bobrowski’s Functional Analysis for Probability and Stochastic Processes.
Theorem. For any , exists and is a.s. unique. Moreover, the map is a Markov operator and restricted to the subspace is the projection onto the subspace .
Proof. First, suppose . Then since the -limit of -measurable functions is again -measurable, we have that is a closed subspace of the Hilbert space . We define the random variable to be the orthogonal projection of onto . Denote the -inner product by . Since for all , taking , for any , we obtain
Note that is a dense subspace of . Thus, if we can show that $Platex X \geq 0$ for any , then by the Markov extension theorem has a unique extension to a Markov operator . Since preserves the continuous linear functional on the dense subset , for any , we then conclude that
Suppose is nonnegative, and that . Since , for some . Since . Since and is nonnegative, we have that
which is a contradiction.