A probability model often contains more information than the question we are ready to use. A gambler sees the first few tosses but not the future; a statistician observes a noisy measurement but not the hidden parameter; a stochastic process is known up to time $t$ but not beyond it. In each case the same problem appears: replace an unknown random variable by the best prediction that is allowed to depend only on the information currently available.

Ordinary expectation answers a question with no information: what single number best summarizes $X$ before anything is observed? Conditional expectation answers the same question after restricting the observer to a chosen collection of observable events. The answer is not usually a number. It is a new random variable, constant on indistinguishable states and measurable with respect to the information being used.

[example: A Prediction That Cannot See the Whole Outcome] Let $\nu$ be the uniform probability measure on $\Omega=\{1,2,3,4,5,6\}$, let $X(k)=k$, and suppose the observer is only told whether the die roll is even or odd. The available information is \begin{align*} \mathcal G=\{\varnothing,\{1,3,5\},\{2,4,6\},\Omega\}. \end{align*} Since the only nontrivial observable atoms are $\{1,3,5\}$ and $\{2,4,6\}$, any $\mathcal G$-measurable predictor must be constant on each of these two sets. On the odd atom, \begin{align*} \nu(\{1,3,5\})&=\frac{3}{6}=\frac{1}{2},\\ \int_{\{1,3,5\}} X\,d\nu &=1\cdot \nu(\{1\})+3\cdot \nu(\{3\})+5\cdot \nu(\{5\})\\ &=1\cdot\frac{1}{6}+3\cdot\frac{1}{6}+5\cdot\frac{1}{6}\\ &=\frac{1+3+5}{6} =\frac{9}{6} =\frac{3}{2}. \end{align*} Therefore the constant value on $\{1,3,5\}$ must be \begin{align*} \frac{\int_{\{1,3,5\}}X\,d\nu}{\nu(\{1,3,5\})} =\frac{3/2}{1/2} =3. \end{align*} Similarly, on the even atom, \begin{align*} \nu(\{2,4,6\})&=\frac{3}{6}=\frac{1}{2},\\ \int_{\{2,4,6\}} X\,d\nu &=2\cdot\frac{1}{6}+4\cdot\frac{1}{6}+6\cdot\frac{1}{6}\\ &=\frac{2+4+6}{6} =\frac{12}{6} =2, \end{align*} so the constant value on $\{2,4,6\}$ must be \begin{align*} \frac{\int_{\{2,4,6\}}X\,d\nu}{\nu(\{2,4,6\})} =\frac{2}{1/2} =4. \end{align*} Thus the conditional expectation of $X$ given parity is the random variable \begin{align*} Y(k)= \begin{cases} 3, & k\in \{1,3,5\},\\ 4, & k\in \{2,4,6\}. \end{cases} \end{align*} It is $\mathcal G$-measurable because the preimages of its possible values are $\{1,3,5\}$ and $\{2,4,6\}$, both of which belong to $\mathcal G$. Finally, $Y$ preserves the averages over every observable event. For the odd atom, \begin{align*} \int_{\{1,3,5\}}Y\,d\nu =3\cdot\nu(\{1,3,5\}) =3\cdot\frac{1}{2} =\frac{3}{2} =\int_{\{1,3,5\}}X\,d\nu, \end{align*} and for the even atom, \begin{align*} \int_{\{2,4,6\}}Y\,d\nu =4\cdot\nu(\{2,4,6\}) =4\cdot\frac{1}{2} =2 =\int_{\{2,4,6\}}X\,d\nu. \end{align*} For $\varnothing$, both integrals are $0$, and for $\Omega$, \begin{align*} \int_\Omega Y\,d\nu &=\int_{\{1,3,5\}}Y\,d\nu+\int_{\{2,4,6\}}Y\,d\nu\\ &=\frac{3}{2}+2 =\frac{7}{2},\\ \int_\Omega X\,d\nu &=\frac{1+2+3+4+5+6}{6} =\frac{21}{6} =\frac{7}{2}. \end{align*} The predictor cannot distinguish individual die rolls inside a parity class, so it replaces each class by its average while preserving the total integral on every event the observer can describe. [/example]

The second condition in the example is the key. The observer cannot distinguish points inside an atom of $\mathcal G$, but the total mass of $X$ over every event the observer can describe must be preserved. Conditional expectation is the unique random variable with exactly those two properties.

## Information and Measurability

Before conditional expectation itself can be defined, the word "information" has to be turned into a mathematical object. The events visible to an observer should be closed under complements and countable unions, because these are the logical operations used to combine observable questions. A smaller collection of events records less information while still supporting those operations.

[definition: Sub-$\sigma$-Algebra] Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space. A sub-$\sigma$-algebra of $\mathcal F$ is a collection $\mathcal G\subset\mathcal F$ such that $(\Omega,\mathcal G)$ is a measurable space. [/definition]

Once the visible events have been isolated, the next question is which random variables can be computed from them. This motivates the measurability condition that turns dependence on information into a precise statement.

[definition: $\mathcal G$-Measurable Random Variable] Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and let $\mathcal G\subset\mathcal F$ be a sub-$\sigma$-algebra. A real-valued random variable $Y:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is $\mathcal G$-measurable if $Y^{-1}(B)\in\mathcal G$ for every Borel set $B\in\mathcal B(\mathbb R)$. [/definition]

These two definitions separate the information itself from the random variables it can observe. With that language in place, the prediction problem becomes a preservation problem: replace $X$ by a $\mathcal G$-measurable random variable without changing integrals over $\mathcal G$-observable events.

## Definition

The main definition states the prediction problem as a preservation problem. We seek a $\mathcal G$-measurable replacement for $X$ whose integrals over all $\mathcal G$-observable events are the same as those of $X$.

[definition: Conditional Expectation Given a Sub-$\sigma$-Algebra] Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, let $\mathcal G\subset\mathcal F$ be a sub-$\sigma$-algebra, and let $X\in L^1(\Omega,\mathcal F,\mathbb P)$, meaning that $X$ is integrable and $\mathcal F$-measurable. A conditional expectation of $X$ given $\mathcal G$ is a real-valued random variable $Y\in L^1(\Omega,\mathcal G,\mathbb P)$ such that \begin{align*} \int_A Y \, d\mathbb P = \int_A X \, d\mathbb P \end{align*} for every $A\in\mathcal G$. [/definition]

The definition asks for a random variable that is visible to the smaller information structure while preserving all integrals over that information. It is not obvious that such a visible representative exists, because the original variable may depend on distinctions that $\mathcal G$ cannot see. The existence question is resolved by treating the integrals of $X$ over $\mathcal G$-events as a measure on the smaller measurable space.

[quotetheorem:1147]

This theorem says that the preservation problem is not merely a useful definition but a well-posed operation. The output is determined only up to almost sure equality, which is exactly the right uniqueness notion for random variables in $L^1$. Conceptually, conditional expectation discards distinctions invisible to $\mathcal G$ while retaining every average that $\mathcal G$ can test. The next step is to connect this sub-$\sigma$-algebra formulation with the more common situation where the available information is generated by an observed random variable.

## Conditioning on Random Variables

An observer often reports a random variable rather than naming a $\sigma$-algebra. To condition on that observation, the theory keeps exactly the events determined by the observed value.

[definition: Conditional Expectation Given a Random Variable] Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, let $X\in L^1(\Omega,\mathcal F,\mathbb P)$, and let $Z:(\Omega,\mathcal F)\to(E,\mathcal E)$ be a random variable. The conditional expectation of $X$ given $Z$ is \begin{align*} \mathbb E[X\mid Z] := \mathbb E[X\mid \sigma(Z)]. \end{align*} [/definition]

After this definition, the phrase “given $Z$” should be read as “given every event determined by $Z$”. This viewpoint is essential when $Z$ is continuous, since events of the form $\{Z=z\}$ often have probability zero and cannot support elementary conditional averages.