The Averaging Characterization of Conditional Expectation

Why Cell Averages Characterize Conditional Expectation

When information is limited to which cell of a finite partition contains an outcome, a conditional expectation must be constant within each cell: it cannot vary according to distinctions the information does not reveal. But constancy alone does not determine its values. The values are fixed by an averaging requirement: within each cell, the proposed conditional expectation must have the same integral as the original random variable.

This tutorial proves that requirement in both directions. The proof is useful because it reduces the defining condition for conditional expectation to a finite collection of equations, one for each cell. It also identifies an important qualification: on a cell of probability zero, the equation cannot determine the constant, so conditional expectations are identified only up to almost-sure equality.

Definition and Finite-Partition Setup

Work on a probability space $(\Omega,\mathcal{F},P)$. Let $X$ be an integrable random variable, meaning $E[|X|]<\infty$. Let $A_1,\ldots,A_m$ be a finite measurable partition of $\Omega$: the cells are pairwise disjoint and their union is $\Omega$. Write $\mathcal{G}=\sigma(A_1,\ldots,A_m)$ for the sigma-field generated by the partition. Its events are exactly the unions of cells.

Definition: A random variable $Z$ is a conditional expectation of $X$ given $\mathcal{G}$, written $Z=E[X\mid\mathcal{G}]$, if (i) $Z$ is integrable, (ii) $Z$ is $\mathcal{G}$-measurable, and (iii) for every $B\in\mathcal{G}$, $$ E[Z\mathbf{1}_B]=E[X\mathbf{1}_B]. $$ Here $\mathbf{1}_B$ is the indicator of $B$. Conditional expectations that differ only on an event of probability zero are regarded as versions of the same conditional expectation.

For this finite partition, suppose $Z$ is constant on each cell, with value $c_i$ on $A_i$. Then $$ E[Z\mathbf{1}_{A_i}]=c_iP(A_i). $$ Thus the condition on the cell $A_i$ is the equation $c_iP(A_i)=E[X\mathbf{1}_{A_i}]$. If $P(A_i)>0$, it determines $c_i$; if $P(A_i)=0$, both sides are zero for integrable $X$, and the equation places no restriction on a finite value assigned to that cell.

The Averaging Characterization

Theorem (Averaging characterization for a finite partition): Let $X$ be integrable, and let $A_1,\ldots,A_m$ be a finite measurable partition. Suppose $Z$ is integrable and constant on each cell. Then $Z=E[X\mid\mathcal{G}]$, where $\mathcal{G}=\sigma(A_1,\ldots,A_m)$, if and only if $$ E[Z\mathbf{1}_{A_i}]=E[X\mathbf{1}_{A_i}] \quad\text{for every }i=1,\ldots,m. $$ Equivalently, if $Z=c_i$ on $A_i$, then $c_iP(A_i)=E[X\mathbf{1}_{A_i}]$ for every cell.

Proof. First suppose $Z=E[X\mid\mathcal{G}]$. Each cell $A_i$ belongs to $\mathcal{G}$. The defining integral condition, applied with $B=A_i$, gives $E[Z\mathbf{1}_{A_i}]=E[X\mathbf{1}_{A_i}]$.

Conversely, suppose the equality holds on every cell. An arbitrary event $B\in\mathcal{G}$ is a union of some of the partition cells. For an index set $I\subseteq\{1,\ldots,m\}$, write $B=\bigcup_{i\in I}A_i$. The cells are disjoint, so $$ E[Z\mathbf{1}_B] =\sum_{i\in I}E[Z\mathbf{1}_{A_i}] =\sum_{i\in I}E[X\mathbf{1}_{A_i}] =E[X\mathbf{1}_B]. $$ The middle equality uses the assumed cell equations; the last follows from disjointness and additivity of the integral. Since $Z$ is integrable and $\mathcal{G}$-measurable by assumption, it satisfies all parts of the definition of $E[X\mid\mathcal{G}]$. $\square$

The theorem also gives the constants directly. On every cell with positive probability, $$ c_i=\frac{E[X\mathbf{1}_{A_i}]}{P(A_i)}. $$ This is the mean of $X$ within that cell. The formula is not used on a zero-probability cell because its denominator is zero; there, any finite constant gives the same integral.

Worked Example: Grouping the Outcomes of a Fair Die

Let $X$ be the outcome of a fair six-sided die. Partition the outcomes into $A_1=\{1,3,5\}$ and $A_2=\{2,4,6\}$. Find the function constant on each cell that is $E[X\mid\mathcal{G}]$, where $\mathcal{G}=\sigma(A_1,A_2)$, and verify the cell equations.

Each cell has probability $3/6=1/2$. On the odd cell, $$ E[X\mathbf{1}_{A_1}] =\frac{1+3+5}{6} =\frac{9}{6} =\frac{3}{2}. $$ Consequently, the required constant is $c_1=(3/2)/(1/2)=3$. Its cell integral is $3P(A_1)=3/2$, as required.

On the even cell, $$ E[X\mathbf{1}_{A_2}] =\frac{2+4+6}{6} =\frac{12}{6} =2. $$ Thus $c_2=2/(1/2)=4$, and $4P(A_2)=2$. The function equal to $3$ on odd outcomes and $4$ on even outcomes matches the integral of $X$ on each cell, so the theorem proves that it is $E[X\mid\mathcal{G}]$.

The events in $\mathcal{G}$ are $\varnothing$, $A_1$, $A_2$, and $\Omega$. The cell equations give the defining equality for $A_1$ and $A_2$; adding them gives it for $\Omega$, and both integrals over $\varnothing$ are zero.

Worked Example: A Continuous Variable Averaged Within Intervals

Let $X\sim\operatorname{Uniform}(0,4)$, and take $A_1=[0,2)$ and $A_2=[2,4]$. Calculate the cell constants. The endpoints have probability zero, so the choice of which cell contains $2$ does not affect the calculation.

The density is $1/4$ on $[0,4]$, and each cell has probability $1/2$. For the first cell, $$ E[X\mathbf{1}_{A_1}] =\int_0^2 \frac{x}{4}\,dx =\left[\frac{x^2}{8}\right]_0^2 =\frac{1}{2}. $$ The corresponding constant is $(1/2)/(1/2)=1$.

For the second cell, $$ E[X\mathbf{1}_{A_2}] =\int_2^4 \frac{x}{4}\,dx =\left[\frac{x^2}{8}\right]_2^4 =\frac{16-4}{8} =\frac{3}{2}. $$ Its constant is $(3/2)/(1/2)=3$. Checking the integrals directly, the candidate function has integral $1(1/2)=1/2$ over $A_1$, and $3(1/2)=3/2$ over $A_2$, exactly matching the integrals of $X$. It is therefore the conditional expectation given the partition.

Uniqueness and Zero-Probability Cells

The cell equations also explain exactly how much freedom there is in choosing a version. On a cell $A_i$ with $P(A_i)>0$, the equation $$ c_iP(A_i)=E[X\mathbf{1}_{A_i}] $$ has exactly one solution for $c_i$. On a cell with probability zero, $E[X\mathbf{1}_{A_i}]=0$, so every finite $c_i$ satisfies the equation. Changing the function on such a cell does not change it except on a null event.

More generally, this proves uniqueness up to almost-sure equality among functions constant on the cells: if two such functions both satisfy the cell equations, their constants agree on every positive-probability cell. They can differ only on zero-probability cells, whose finite union still has probability zero.

Worked Example: A Null Cell Does Not Determine the Value

Let $X\sim\operatorname{Uniform}(0,1)$, and partition $[0,1]$ into $A_1=\{1/2\}$ and $A_2=[0,1]\setminus\{1/2\}$. Define $Z$ to be $0$ on $A_1$ and $1/2$ on $A_2$. Verify the cell equations and explain why the value at $1/2$ can be changed.

The singleton $A_1$ has probability zero. Therefore $E[X\mathbf{1}_{A_1}]=0$, and $E[Z\mathbf{1}_{A_1}]=0$. For the other cell, removing one point does not change the integral, so $$ E[X\mathbf{1}_{A_2}]=\int_0^1 x\,dx=\frac{1}{2}. $$ Since $Z=1/2$ on $A_2$ and $P(A_2)=1$, $$ E[Z\mathbf{1}_{A_2}]=\frac{1}{2}P(A_2)=\frac{1}{2}. $$ Both cell equations hold. If the value assigned to $Z$ at $1/2$ were changed to any other finite number, its integral on that singleton would still be zero. The resulting function would be another version of the same conditional expectation.

Common Mistake: Matching Values Instead of Integrals

A tempting but incorrect test is to require the conditional expectation to equal $X$ at each outcome, or to check only that its overall expectation equals $E[X]$. The first condition is generally too strong: the function must be constant on each cell even when $X$ varies within that cell. The second condition checks only the union of all cells, not each cell separately.

For instance, in the fair-die example, the function constantly equal to $7/2$ has the same overall expectation as $X$, since $E[X]=7/2$. But on the odd cell its integral is $(7/2)(1/2)=7/4$, whereas $E[X\mathbf{1}_{A_1}]=3/2$. The cell equation fails, so this constant function is not the conditional expectation given the odd/even partition.

The correct check is cell by cell: multiply the proposed constant on each cell by that cell’s probability and compare the result with the integral of $X$ over that cell. Matching all cell equations is both necessary and sufficient.

Key takeaway: For a finite partition, a cell-constant, integrable function is a conditional expectation precisely when it matches the original variable’s integral on every cell. Positive-probability cells determine the constants; zero-probability cells do not.

Check Your Understanding

Use the cell equations to answer each question.

In the finite-partition theorem, why does matching the integrals on the cells imply matching the integrals on every event in the generated sigma-field?
If $Z=c_i$ on a cell $A_i$ with $P(A_i)>0$, solve the cell equation for $c_i$.
Suppose a partition has a cell of probability zero. What does the cell equation say about the value of a cell-constant conditional expectation there?
For a fair die partitioned into $\{1,2\}$ and $\{3,4,5,6\}$, calculate the conditional-expectation constant on each cell.
Why is equality of overall expectations alone not enough to verify a candidate conditional expectation for a finite partition?