The Tower Property of Conditional Expectation

Why Conditional Expectations Can Be Taken in Stages

Conditional expectation replaces a random variable by its average relative to specified information. If one first averages using a finer sigma-algebra and then averages using a coarser one, the result should agree with averaging directly using the coarser information. The tower property makes this precise. Its proof depends on two facts: the defining integral identities for conditional expectation, and the inclusion of the coarser sigma-algebra in the finer one.

Throughout, \((\Omega,\mathcal{F},\mathbb{P})\) is a probability space, and \(X\) is a real-valued integrable random variable, meaning \(\mathbb{E}[|X|]<\infty\). For a sub-sigma-algebra \(\mathcal{A}\subseteq\mathcal{F}\), a conditional expectation \(\mathbb{E}[X\mid\mathcal{A}]\) is an integrable, \(\mathcal{A}\)-measurable random variable \(Y\) such that \(\mathbb{E}[Y\chi_A]=\mathbb{E}[X\chi_A]\) for every \(A\in\mathcal{A}\). Here \(\chi_A\) is the indicator of \(A\). Conditional expectations are determined only up to equality \(\mathbb{P}\)-almost surely; any particular measurable representative may be used.

Proof. Choose versions \(U=\mathbb{E}[X\mid\mathcal{H}]\) and \(V=\mathbb{E}[U\mid\mathcal{G}]\). By the definition of conditional expectation, \(U\) is integrable, so it is a valid input for conditioning on \(\mathcal{G}\). The variable \(V\) is integrable and \(\mathcal{G}\)-measurable. For any \(A\in\mathcal{G}\), nesting gives \(A\in\mathcal{H}\), and the two defining identities give \(\mathbb{E}[V\chi_A]=\mathbb{E}[U\chi_A]=\mathbb{E}[X\chi_A]\). Thus \(V\) satisfies the defining identities for \(\mathbb{E}[X\mid\mathcal{G}]\).

For completeness, these identities identify an integrable \(\mathcal{G}\)-measurable variable uniquely up to a null set. Indeed, if \(V\) and \(W\) both satisfy them, then \(D=V-W\) is integrable and \(\mathcal{G}\)-measurable, and \(\mathbb{E}[D\chi_A]=0\) for every \(A\in\mathcal{G}\). For each positive integer \(n\), the event \(A_n=\{D>1/n\}\) belongs to \(\mathcal{G}\), so \(0=\mathbb{E}[D\chi_{A_n}]\ge (1/n)\mathbb{P}(A_n)\). Hence every \(A_n\) is null, and \(\{D>0\}=\bigcup_{n\ge1}A_n\) is null. Applying the same argument to \(-D\) shows that \(\{D<0\}\) is null. Therefore \(V=W\) almost surely. Applying this uniqueness to \(V\) and \(\mathbb{E}[X\mid\mathcal{G}]\) proves the assertion. \(\square\)

Conditional Averages on Finite Partitions

A finite partition of \(\Omega\) is a finite collection of disjoint measurable sets whose union is \(\Omega\). Its cells generate a sigma-algebra consisting of unions of cells. On each cell with positive probability, conditioning on that sigma-algebra gives the probability-weighted average of \(X\) over the cell. Cells of probability zero do not affect the conditional expectation and can be assigned arbitrary finite values.

Proof. Define \(Y\) to have the displayed constant value on each positive-probability cell, and assign value \(0\) on each zero-probability cell. It is \(\mathcal{A}\)-measurable. It is integrable because \(\sum_{\mathbb{P}(C_j)>0}\mathbb{E}[|Y|\chi_{C_j}] =\sum_{\mathbb{P}(C_j)>0}|\mathbb{E}[X\chi_{C_j}]| \le\sum_{j=1}^k\mathbb{E}[|X|\chi_{C_j}] =\mathbb{E}[|X|]<\infty\). For every positive-probability cell, the definition gives \(\mathbb{E}[Y\chi_{C_j}]=\mathbb{E}[X\chi_{C_j}]\). For a zero-probability cell both integrals are zero. Every event in \(\mathcal{A}\) is a union of cells, so adding these identities over its cells gives the defining identity for every event in \(\mathcal{A}\). Thus \(Y\) is a version of the conditional expectation. \(\square\)

What the Intermediate Conditional Expectation Need Not Be

The tower property does not say that \(\mathbb{E}[X\mid\mathcal{H}]\) is \(\mathcal{G}\)-measurable. The inclusion \(\mathcal{G}\subseteq\mathcal{H}\) implies that every \(\mathcal{G}\)-measurable variable is also \(\mathcal{H}\)-measurable, not the reverse. For the iterated conditional expectation to be defined, the input \(\mathbb{E}[X\mid\mathcal{H}]\) only needs to be integrable, which it is by definition.

Why the Nesting Condition Matters

The proof uses nesting at a specific point: to apply the defining identity for conditioning on \(\mathcal{H}\) to every event in \(\mathcal{G}\). Without \(\mathcal{G}\subseteq\mathcal{H}\), the argument does not establish the tower property, and the conclusion can fail.

Common mistake. A tempting but incorrect justification is: “Since \(\mathcal{G}\subseteq\mathcal{H}\), the \(\mathcal{H}\)-measurable variable \(U=\mathbb{E}[X\mid\mathcal{H}]\) is \(\mathcal{G}\)-measurable.” The inclusion goes in the opposite direction for measurability, as the preceding example shows. The correction is to note that \(U\) is integrable, so \(\mathbb{E}[U\mid\mathcal{G}]\) is defined; then for each \(A\in\mathcal{G}\), use \(A\in\mathcal{H}\) to obtain \(\mathbb{E}[\mathbb{E}[U\mid\mathcal{G}]\chi_A]=\mathbb{E}[U\chi_A]=\mathbb{E}[X\chi_A]\). This identifies the result by the defining property and almost-sure uniqueness.

Proof. Apply the defining identity for conditional expectation to \(A=\Omega\), which belongs to \(\mathcal{G}\). Since \(\chi_\Omega=1\), it gives \(\mathbb{E}[\mathbb{E}[X\mid\mathcal{G}]]=\mathbb{E}[X]\). \(\square\)