Fixed Points of Conditional Expectation

Conditioning as an Operator That Retains Certain Variables

Conditional expectation replaces a random variable by a version that uses only the information in a chosen sigma-algebra. Some variables already use only that information, so conditioning leaves them unchanged, at least almost surely. Characterizing exactly which variables have this property gives a useful way to understand conditional expectation as an operator.

Fix a probability space \((\Omega,\mathcal F,\mathbb P)\) and a sub-sigma-algebra \(\mathcal G\subseteq\mathcal F\). For \(X\in L^1(\mathcal F)\), write \(\mathbb E[X\mid\mathcal G]\) for a conditional expectation of \(X\) given \(\mathcal G\). It is a \(\mathcal G\)-measurable integrable random variable whose integral over every event in \(\mathcal G\) agrees with that of \(X\). Such a variable is unique up to almost-sure equality.

The defining integral condition already suggests the central fact: when \(X\) is itself measurable with respect to \(\mathcal G\), it meets the definition of its own conditional expectation. The converse is true in the almost-sure sense: if conditioning leaves \(X\) unchanged, then \(X\) has a \(\mathcal G\)-measurable version.

Fixed Points and Measurability

Definition: An integrable random variable \(X\) is a fixed point of conditioning on \(\mathcal G\) if \(\mathbb E[X\mid\mathcal G]=X\) almost surely. This means equality of the random variables outside a set of probability zero; it need not mean pointwise equality at every \(\omega\in\Omega\).

The qualification “almost surely” matters. Conditional expectations are defined only up to almost-sure equality, and integrable random variables are often treated as equivalence classes under that relation. The fixed-point characterization is therefore naturally a statement about equivalence classes.

Theorem: For \(X\in L^1(\mathcal F)\), the following are equivalent:

\(\mathbb E[X\mid\mathcal G]=X\) almost surely;
\(X\) has a \(\mathcal G\)-measurable version.

In particular, if \(X\) is \(\mathcal G\)-measurable, then \(\mathbb E[X\mid\mathcal G]=X\) almost surely.

Proof. Suppose first that \(\mathbb E[X\mid\mathcal G]=X\) almost surely. A conditional expectation has a \(\mathcal G\)-measurable version by definition. Since that version equals \(X\) almost surely, \(X\) has a \(\mathcal G\)-measurable version. Conversely, suppose \(Y\) is a \(\mathcal G\)-measurable version of \(X\). Then \(Y\) is integrable and \(X=Y\) almost surely. For every \(A\in\mathcal G\), \[ \int_A X\,d\mathbb P=\int_A Y\,d\mathbb P, \] because changing an integrable random variable on a null set does not change its integral. Since \(Y\) is \(\mathcal G\)-measurable, it satisfies the defining integral condition for \(\mathbb E[Y\mid\mathcal G]\); hence \(\mathbb E[Y\mid\mathcal G]=Y\) almost surely. The displayed equality of integrals shows that \(Y\) also satisfies the defining condition for \(\mathbb E[X\mid\mathcal G]\). By almost-sure uniqueness, \[ \mathbb E[X\mid\mathcal G]=Y=X\quad\text{almost surely}. \] This proves both directions. \(\square\)

Idempotence and the Range

Define the conditional expectation operator \(T_{\mathcal G}\) on \(L^1(\mathcal F)\) by \[ T_{\mathcal G}X=\mathbb E[X\mid\mathcal G]. \] Because conditional expectations are unique up to almost-sure equality, this defines an operator on equivalence classes. Its values are integrable and \(\mathcal G\)-measurable, so applying the same conditioning operation again acts on a variable already measurable with respect to \(\mathcal G\).

Theorem: Conditional expectation onto \(\mathcal G\) is idempotent: \[ T_{\mathcal G}(T_{\mathcal G}X)=T_{\mathcal G}X \quad\text{for every }X\in L^1(\mathcal F). \] Its range is exactly \(L^1(\mathcal G)\), viewed as a subspace of \(L^1(\mathcal F)\). Equivalently, the range consists of the almost-sure equivalence classes that have \(\mathcal G\)-measurable integrable representatives.

Proof. Let \(X\in L^1(\mathcal F)\), and choose a \(\mathcal G\)-measurable version \(Y\) of \(T_{\mathcal G}X\). The variable \(Y\) is integrable. Since it is \(\mathcal G\)-measurable, the fixed-point characterization gives \[ T_{\mathcal G}Y=Y\quad\text{almost surely}. \] Since \(Y=T_{\mathcal G}X\) almost surely and conditional expectation depends only on the almost-sure equivalence class of its input, \[ T_{\mathcal G}(T_{\mathcal G}X)=T_{\mathcal G}Y=Y=T_{\mathcal G}X \quad\text{almost surely}. \] This proves idempotence. For the range, every value \(T_{\mathcal G}X\) has a \(\mathcal G\)-measurable integrable version, so the range is contained in \(L^1(\mathcal G)\). Conversely, if \(Y\in L^1(\mathcal G)\), the fixed-point characterization gives \(T_{\mathcal G}Y=Y\) almost surely. Thus every element of \(L^1(\mathcal G)\) is in the range. The two inclusions prove the range statement. \(\square\)

An idempotent operator acts as the identity on its range. Here, conditioning first maps an arbitrary integrable variable into the class of \(\mathcal G\)-measurable variables; conditioning again cannot change that class. The operator is therefore a projection in the algebraic sense. This description does not assert that it is an orthogonal projection in \(L^1\); the Hilbert-space projection interpretation belongs to \(L^2\).

Worked Examples

Worked Example: Averaging on a Finite Partition

Let \(\Omega=\{1,2,3,4\}\), with each point having probability \(1/4\), and let \(\mathcal G\) be generated by the partition \(\{\{1,2\},\{3,4\}\}\). On each atom, conditional expectation is the average of the values there. For \(X=(1,5,2,10)\), where the coordinates list the values at \(1,2,3,4\),

\mathbb E[X\mid\mathcal G] = \left(\frac{1+5}{2},\frac{1+5}{2}, \frac{2+10}{2},\frac{2+10}{2}\right) =(3,3,6,6).

This result is not a fixed point equal to \(X\): for example, at point \(1\) the values are \(3\) and \(1\), and every point has positive probability. By contrast, \(Y=(3,3,6,6)\) is \(\mathcal G\)-measurable because it is constant on each atom. Its conditional averages are \(3\) on \(\{1,2\}\) and \(6\) on \(\{3,4\}\), so \(\mathbb E[Y\mid\mathcal G]=Y\). In this finite setting the fixed points are precisely the variables constant on each partition atom.

Worked Example: Conditioning on the Trivial Sigma-Algebra

Suppose \(\mathcal G=\{\varnothing,\Omega\}\), and \(\Omega=\{1,2,3,4\}\) has the uniform probability distribution. A \(\mathcal G\)-measurable random variable must be constant on \(\Omega\). If \(X=(-1,0,2,5)\), its conditional expectation is the constant equal to its mean:

\mathbb E[X\mid\mathcal G] =\mathbb E[X] =\frac{-1+0+2+5}{4} =\frac{3}{2}.

To check the defining condition, the only events in \(\mathcal G\) are \(\varnothing\) and \(\Omega\). Both variables have integral zero on \(\varnothing\), while \[ \int_\Omega X\,d\mathbb P=\frac{3}{2} =\int_\Omega \frac{3}{2}\,d\mathbb P. \] Thus the constant \(3/2\) is indeed a conditional expectation. The fixed points are exactly the variables that are constant almost surely: conditioning on the trivial sigma-algebra retains only the mean.

Worked Example: Conditioning on One Coordinate

Take \(\Omega=[0,1]^2\) with product Lebesgue probability measure. Write \(U(u,v)=u\) and \(V(u,v)=v\), and let \(\mathcal G=\sigma(U)\). For \(X=U+V\), consider \(Y=U+1/2\), which is \(\mathcal G\)-measurable. Each event in \(\mathcal G\) has the form \(B\times[0,1]\) for a Borel set \(B\subseteq[0,1]\). Fubini's theorem gives

\int_{B\times[0,1]}X\,d\mathbb P =\int_B\int_0^1(u+v)\,dv\,du =\int_B\left(u+\frac12\right)du =\int_{B\times[0,1]}Y\,d\mathbb P.

Therefore \(Y\) satisfies the defining integral condition, and \(\mathbb E[X\mid\mathcal G]=U+1/2\) almost surely. The variable \(U\) itself is fixed by conditioning, since it is \(\mathcal G\)-measurable. In contrast, \(X\) is not fixed: \(U+V\) differs from \(U+1/2\) whenever \(V\ne1/2\), an event of probability one. This example shows how conditioning on one coordinate averages out the other coordinate.

Almost-Sure Equality Is Not Pointwise Measurability

A common source of confusion is to infer that a random variable must itself be \(\mathcal G\)-measurable whenever it is fixed almost surely. The theorem gives a \(\mathcal G\)-measurable version, not necessarily pointwise measurability of the original representative.

For a concrete example, let \(\Omega=[0,1]\) with its Borel sigma-algebra and Lebesgue probability measure, let \(\mathcal G=\{\varnothing,\Omega\}\), and let \(N=\{0\}\). The indicator \(X=\mathbf 1_N\) is Borel-measurable and equals zero almost surely. Hence \(\mathbb E[X\mid\mathcal G]=0=X\) almost surely. But \(X\) is not \(\mathcal G\)-measurable: it is not constant at every point. It does have the \(\mathcal G\)-measurable version \(Y=0\). Working with equivalence classes, or consistently saying “up to almost-sure equality,” resolves the distinction.

Why the Characterization Matters

The range description gives a direct interpretation of conditional expectation: it is an operator that takes integrable variables into the space of variables measurable with respect to the available information, and it fixes every element of that space. Idempotence is a concise algebraic expression of this idea. Once the output depends only on \(\mathcal G\), repeating the same conditioning adds no further change.

The exact formulation also prevents two errors. First, being fixed does not mean that a random variable has a particular constant value; it means that the variable is measurable with respect to the chosen sigma-algebra, up to null sets. Second, equality with a conditional expectation is naturally almost sure, not necessarily pointwise. Keeping these qualifications visible makes the fixed-point and range statements correct even when the sigma-algebra is not complete.

Check Your Understanding

Use the fixed-point characterization and the defining integral condition to answer the following questions.

What does it mean for an integrable random variable to be a fixed point of conditioning on \(\mathcal G\)?
Why does every \(\mathcal G\)-measurable integrable variable belong to the range of \(T_{\mathcal G}\)?
How does the fixed-point characterization imply idempotence?
For a finite partition, what condition on the values of a variable makes it a fixed point?
Why might a variable fixed almost surely by conditioning fail to be pointwise \(\mathcal G\)-measurable?