Independence Removes Information About the Mean
Conditional expectation records the part of an integrable random variable that is determined by a specified sigma-algebra. If that sigma-algebra is independent of the random variable, it contains no information about the variable’s values. The conditional expectation is then constant: it equals the ordinary expectation. The essential point is that independence must hold between the sigma-algebra and the random variable itself, not merely between the sigma-algebra and some other variable used to construct it.
Throughout, let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let \(\mathcal{G}\subseteq\mathcal{F}\) be a sigma-algebra, and let \(X:\Omega\to\mathbb{R}\) be a measurable, integrable random variable. Integrability means \(\mathbb{E}[|X|]<\infty\). All equalities between conditional expectations are understood \(\mathbb{P}\)-almost surely.
The event-integral identity in the definition is the useful test. To show that a candidate \(Z\) is the conditional expectation, it suffices to check its measurability and integrability, then verify the identity for every event in \(\mathcal{G}\). In the independent case, the constant \(\mathbb{E}[X]\) passes this test.
The Independence Theorem
Proof. Fix \(B\in\mathcal{G}\). First suppose \(U\) is a nonnegative, simple, \(\sigma(X)\)-measurable random variable, written using disjoint sets as \(U=\sum_{j=1}^k a_j\chi_{A_j}\), where \(a_j\ge 0\) and \(A_j\in\sigma(X)\). Independence gives \(\mathbb{P}(A_j\cap B)=\mathbb{P}(A_j)\mathbb{P}(B)\) for each \(j\). Therefore, \[ \mathbb{E}[U\chi_B] =\sum_{j=1}^k a_j\mathbb{P}(A_j\cap B) =\left(\sum_{j=1}^k a_j\mathbb{P}(A_j)\right)\mathbb{P}(B) =\mathbb{E}[U]\mathbb{P}(B). \] By the definition of the Lebesgue integral, a nonnegative \(\sigma(X)\)-measurable random variable is the increasing pointwise limit of nonnegative simple \(\sigma(X)\)-measurable random variables. Apply the displayed identity to such an approximation of \(X^+\). The Monotone Convergence Theorem applies on both sides, since the approximants and their products with \(\chi_B\) are nonnegative and increasing. It follows that \(\mathbb{E}[X^+\chi_B]=\mathbb{E}[X^+]\mathbb{P}(B)\). Applying the same argument to \(X^-\) gives \(\mathbb{E}[X^-\chi_B]=\mathbb{E}[X^-]\mathbb{P}(B)\). Integrability ensures both expectations are finite, so subtraction is valid and yields \[ \int_B X\,d\mathbb{P} =\mathbb{E}[X\chi_B] =\big(\mathbb{E}[X^+]-\mathbb{E}[X^-]\big)\mathbb{P}(B) =\mathbb{E}[X]\mathbb{P}(B). \] The constant random variable \(c=\mathbb{E}[X]\) is integrable and \(\mathcal{G}\)-measurable, and \(\int_B c\,d\mathbb{P}=c\mathbb{P}(B)\). Thus it satisfies the defining event-integral identity for \(\mathbb{E}[X\mid\mathcal{G}]\). For completeness, uniqueness follows from that identity: if two integrable \(\mathcal{G}\)-measurable candidates differ by \(D\), then \(\int_B D\,d\mathbb{P}=0\) for all \(B\in\mathcal{G}\). For every positive integer \(k\), take \(B=\{D\ge 1/k\}\in\mathcal{G}\). The integral identity implies \(0=\int_B D\,d\mathbb{P}\ge \mathbb{P}(B)/k\), so \(\mathbb{P}(D\ge 1/k)=0\). Taking \(B=\{D\le -1/k\}\) gives \(\mathbb{P}(D\le-1/k)=0\). The countable unions of these sets cover \(\{D\ne0\}\), up to any null set on which versions are not finite. Hence \(D=0\) almost surely, proving uniqueness and the theorem. \(\square\)
The probability space here is finite in total measure, since \(\mathbb{P}(\Omega)=1\). No assumption that \(\mathcal{G}\) is generated by finitely many events is needed. The proof uses independence for every event in \(\mathcal{G}\), and integrability to ensure that the positive and negative parts have finite expectations.
Worked Example: A Coin Outcome Independent of a Die
Let \(\Omega=\{H,T\}\times\{1,2,3,4,5,6\}\), with each of the twelve outcomes having probability \(1/12\). Let \(N\) be the die result, let \(X=2\) on outcomes with \(H\) and \(X=0\) on outcomes with \(T\), and set \(\mathcal{G}=\sigma(N)\). Compute \(\mathbb{E}[X\mid\mathcal{G}]\).
The event \(\{X=2\}\) is independent of every event determined by the die: for any subset \(C\subseteq\{1,2,3,4,5,6\}\), the event \(\{N\in C\}\) contains \(|C|\) outcomes with \(H\), so \(\mathbb{P}(X=2,N\in C)=|C|/12=(1/2)(|C|/6)=\mathbb{P}(X=2)\mathbb{P}(N\in C)\). The same factorization holds for \(\{X=0\}\), which is the complement of \(\{X=2\}\). These are the only possible level events of \(X\), so \(\sigma(X)\) and \(\mathcal{G}\) are independent.
Directly, \(\mathbb{E}[X]=2(1/2)+0(1/2)=1\). For each \(C\subseteq\{1,2,3,4,5,6\}\), the event-integral test gives \(\int_{\{N\in C\}}X\,d\mathbb{P}=2|C|/12=|C|/6\), while \(\int_{\{N\in C\}}1\,d\mathbb{P}=\mathbb{P}(N\in C)=|C|/6\). Every event in \(\mathcal{G}\) has this form, so the constant \(1\) satisfies the defining identity. Check: for \(C=\{2,5\}\), the integral is \(2(2/12)=1/3\), and the constant’s integral is \(1(2/6)=1/3\), as required. Therefore \(\mathbb{E}[X\mid\mathcal{G}]=1\) almost surely.
Worked Example: Testing Against an Independent Event
Let \(X\) take the value \(4\) with probability \(1/4\) and \(0\) with probability \(3/4\). Let \(C\) be an event independent of \(X\) with \(\mathbb{P}(C)=1/3\), and put \(\mathcal{G}=\sigma(C)\). Find the conditional expectation.
The expectation is \(\mathbb{E}[X]=4(1/4)+0(3/4)=1\). Since \(C\) is independent of \(X\), \(\sigma(C)\) is independent of \(\sigma(X)\): each event in \(\sigma(C)=\{\varnothing,C,C^c,\Omega\}\) factors with every event determined by \(X\). The theorem therefore gives \(\mathbb{E}[X\mid\mathcal{G}]=1\) almost surely.
For an explicit check, independence gives \(\mathbb{P}(X=4,C)=(1/4)(1/3)=1/12\). Thus \(\int_C X\,d\mathbb{P}=4(1/12)=1/3\), which equals \(\int_C1\,d\mathbb{P}=\mathbb{P}(C)=1/3\). Also, \(\mathbb{P}(C^c)=2/3\) and \(\mathbb{P}(X=4,C^c)=(1/4)(2/3)=1/6\), so \(\int_{C^c}X\,d\mathbb{P}=4(1/6)=2/3=\int_{C^c}1\,d\mathbb{P}\). The identity holds on \(\varnothing\) and \(\Omega\) as well: both integrals are \(0\) on the empty event and \(1\) on \(\Omega\). These are all the events in \(\mathcal{G}\). Check: the two nontrivial event integrals sum to \(1/3+2/3=1=\mathbb{E}[X]\).
Independence of a Component Is Not Enough
A common error is to infer that a variable is independent of \(\mathcal{G}\) because one component in its construction is independent of \(\mathcal{G}\). The theorem requires independence of \(\sigma(X)\) and \(\mathcal{G}\) for the full random variable \(X\). A sign can be independent of a magnitude while the signed product still reveals information about that magnitude.
Worked Example: An Independent Sign with a Dependent Product
Let \(\Omega=\{(n,s):n\ge1,\ s\in\{-1,1\}\}\), with \(\mathbb{P}(\{(n,s)\})=2^{-n-1}\). The total mass is \(\sum_{n\ge1}2(2^{-n-1})=\sum_{n\ge1}2^{-n}=1\). Define \(N(n,s)=n\), \(S(n,s)=s\), \(X=NS\), and \(\mathcal{G}=\sigma(N)\).
For each \(n\), the two points with \(N=n\) have probabilities \(2^{-n-1}\) each. Hence \(\mathbb{P}(N=n)=2^{-n}\), \(\mathbb{P}(S=1)=\mathbb{P}(S=-1)=1/2\), and \(\mathbb{P}(N=n,S=s)=2^{-n-1}=\mathbb{P}(N=n)\mathbb{P}(S=s)\). Summing this equality over any set of values of \(N\) shows that \(S\) is independent of \(\mathcal{G}\).
However, \(X\) is not independent of \(\mathcal{G}\). Take \(A=\{|X|=1\}=\{N=1\}\in\sigma(X)\) and \(B=\{N=1\}\in\mathcal{G}\). Then \(\mathbb{P}(A\cap B)=1/2\), whereas \(\mathbb{P}(A)\mathbb{P}(B)=(1/2)(1/2)=1/4\). The independence theorem cannot be applied to \(X\).
Nevertheless, \(\mathbb{E}[X\mid\mathcal{G}]=0\). First, \(\mathbb{E}[|X|]=\sum_{n\ge1}n2^{-n}=2\), so \(X\) is integrable. For each atom \(\{N=n\}\), \(\int_{\{N=n\}}X\,d\mathbb{P}=n2^{-n-1}-n2^{-n-1}=0\). Every event \(B\in\mathcal{G}\) is a union of such atoms. The sum of the atom integrals is valid because \(\int|X|\,d\mathbb{P}<\infty\), and it gives \(\int_BX\,d\mathbb{P}=0=\int_B0\,d\mathbb{P}\). The zero random variable is integrable and \(\mathcal{G}\)-measurable, so the defining identity and uniqueness give the claim. Check: on \(\{N=1\}\), the two values of \(X\) are \(1\) and \(-1\), each with probability \(1/4\); their contributions are \(1(1/4)+(-1)(1/4)=0\), agreeing with the atom calculation.
Using the Result Correctly
When applying the theorem, first identify the exact random variable \(X\) and the sigma-algebra \(\mathcal{G}\). Then check integrability and verify independence between \(\sigma(X)\) and \(\mathcal{G}\). Only after those checks can the conditional expectation be replaced by the constant \(\mathbb{E}[X]\). In particular, knowing that a sign is independent of some information does not establish that a product involving the sign is independent of that information.
The test-event formulation also explains why the conclusion is an almost-sure equality rather than a pointwise statement about every possible version: conditional expectations are specified by their integrals over \(\mathcal{G}\)-events, and the uniqueness result identifies them only up to null sets. Independence makes the constant pass every one of those tests.
Check Your Understanding
For each question, identify the sigma-algebra and check the required hypotheses before using the theorem.
- If \(X\) is integrable and independent of \(\mathcal{G}\), what event-integral identity verifies that the constant \(\mathbb{E}[X]\) is a conditional expectation?
- In the finite coin-and-die example, calculate \(\int_{\{N\in\{1,4,6\}\}}X\,d\mathbb{P}\) and compare it with the integral of the proposed conditional expectation over the same event.
- For the independent-event example, compute \(\int_{C^c}X\,d\mathbb{P}\) and verify the probability used in the calculation.
- In the signed-product example, exhibit events in \(\sigma(X)\) and \(\mathcal{G}\) that disprove independence.
- Why does independence of \(S\) and \(\mathcal{G}\) not, by itself, allow the theorem to be applied to \(X=NS\)?