What the Event Integrals Determine
Conditional expectation is not generally determined by a formula evaluated at each outcome. Its defining information is instead a collection of integral identities: for every event visible to the conditioning sigma-algebra, the candidate has the same integral as the original random variable. These identities determine the candidate up to changes on a probability-zero set, but they need not determine its value at every individual outcome.
Throughout, let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, so \(\mathbb{P}(\Omega)=1\), and let \(\mathcal{G}\subseteq\mathcal{F}\) be a sigma-algebra. The probability space is finite as a measure space; it is not assumed complete. Let \(X:\Omega\to\mathbb{R}\) be an integrable random variable, meaning \(\int_\Omega |X|\,d\mathbb{P}<\infty\). Candidate conditional expectations are real-valued, integrable, and \(\mathcal{G}\)-measurable. Statements that two such candidates are equal almost surely mean that the set where they differ has \(\mathbb{P}\)-measure zero.
The event condition is stronger than matching the unconditional expectation: taking \(A=\Omega\) gives \(\int_\Omega Y\,d\mathbb{P}=\int_\Omega X\,d\mathbb{P}\), but the definition requires the equality for every event in \(\mathcal{G}\). All the integrals in this definition are finite because \(X\) and \(Y\) are integrable.
From the Radon–Nikodym Construction to the Characterization
The construction using the Radon–Nikodym theorem gives a direct reason for the event identities. Write \(\mathbb{P}_{\mathcal{G}}\) for the restriction of \(\mathbb{P}\) to \(\mathcal{G}\), and define, for \(A\in\mathcal{G}\), $$ \nu(A)=\int_A X\,d\mathbb{P}. $$ Because \(X\) is integrable, \(\nu\) is a finite signed measure on \((\Omega,\mathcal{G})\). If \(\mathbb{P}_{\mathcal{G}}(A)=0\), then \(\int_A X\,d\mathbb{P}=0\), so \(\nu\ll\mathbb{P}_{\mathcal{G}}\). The Radon–Nikodym theorem therefore supplies an integrable \(\mathcal{G}\)-measurable derivative \(Y=d\nu/d\mathbb{P}_{\mathcal{G}}\), characterized by its integrals over sets in \(\mathcal{G}\).
Proof. By the defining property of the Radon–Nikodym derivative, for each \(A\in\mathcal{G}\), $$ \int_A Y\,d\mathbb{P}=\int_A Y\,d\mathbb{P}_{\mathcal{G}}=\nu(A)=\int_A X\,d\mathbb{P}. $$ Thus the constructed derivative satisfies the event identities. Now let \(Z\) be any other integrable, \(\mathcal{G}\)-measurable candidate satisfying them. Subtracting the identities for \(Y\) and \(Z\), using linearity of the integral for integrable functions, gives $$ \int_A (Y-Z)\,d\mathbb{P}=0\qquad\text{for every }A\in\mathcal{G}. $$ We prove that this forces \(Y=Z\) almost surely. Put \(H=Y-Z\), which is integrable and \(\mathcal{G}\)-measurable. For each integer \(n\ge1\), the set \(A_n=\{H\ge1/n\}\) belongs to \(\mathcal{G}\). Since \(H\ge1/n\) on \(A_n\), monotonicity of the integral gives $$ 0=\int_{A_n}H\,d\mathbb{P}\ge\frac{1}{n}\mathbb{P}(A_n). $$ Therefore \(\mathbb{P}(A_n)=0\). Every point where \(H>0\) belongs to some \(A_n\), so \(\{H>0\}=\bigcup_{n\ge1}A_n\) is null. Similarly, \(B_n=\{H\le-1/n\}\in\mathcal{G}\) and $$ 0=\int_{B_n}H\,d\mathbb{P}\le-\frac{1}{n}\mathbb{P}(B_n), $$ so each \(B_n\) is null and \(\{H<0\}=\bigcup_{n\ge1}B_n\) is null. Hence \(\{Y\ne Z\}=\{H>0\}\cup\{H<0\}\) is null, proving \(Y=Z\) almost surely. \(\square\)
The proof uses only integrability, measurability with respect to \(\mathcal{G}\), and the identities on every event in \(\mathcal{G}\). In particular, it does not require the probability space to be complete. The sets used to detect positive or negative differences are in \(\mathcal{G}\), which is exactly what allows the event identities to be applied.
Worked Example: Conditional Expectation on a Finite Partition
Let \(\Omega=\{1,2,3,4\}\), let \(\mathcal{F}\) be its power set, and give each point probability \(1/4\). Let \(\mathcal{G}\) be the sigma-algebra generated by the partition \(C_1=\{1,2\}\), \(C_2=\{3,4\}\). Suppose \(X(1)=2\), \(X(2)=0\), \(X(3)=1\), and \(X(4)=5\). Find and verify a conditional expectation.
Every \(\mathcal{G}\)-measurable function is constant on each partition cell. On \(C_1\), the integral of \(X\) is \((2+0)/4=1/2\), while \(\mathbb{P}(C_1)=2/4=1/2\), so the matching constant is \((1/2)/(1/2)=1\). On \(C_2\), the integral of \(X\) is \((1+5)/4=6/4=3/2\), and \(\mathbb{P}(C_2)=2/4=1/2\), so the matching constant is \((3/2)/(1/2)=3\). Define \(Y=1\) on \(C_1\) and \(Y=3\) on \(C_2\).
The events in \(\mathcal{G}\) are exactly \(\varnothing,C_1,C_2,\Omega\). On \(C_1\), \(\int_{C_1}Y\,d\mathbb{P}=1\cdot(1/2)=1/2=\int_{C_1}X\,d\mathbb{P}\). On \(C_2\), \(\int_{C_2}Y\,d\mathbb{P}=3\cdot(1/2)=3/2=\int_{C_2}X\,d\mathbb{P}\). On \(\Omega\), both integrals equal \(1/2+3/2=2\); on \(\varnothing\), both equal zero. Thus the identities hold for every conditioning event. Check: the values of \(Y\) are the averages of \(X\) on the two cells: \((2+0)/2=1\) and \((1+5)/2=3\).
Worked Example: Conditioning on a Trivial Sigma-Algebra
Let \(\Omega=[0,1]\), \(\mathcal{F}=\mathcal{B}_{[0,1]}\), and let \(\mathbb{P}\) be Lebesgue probability measure on this interval. Take \(X(x)=x\) and \(\mathcal{G}=\{\varnothing,\Omega\}\). Show that the constant \(Y=1/2\) satisfies the event-integral condition.
The function \(X\) is real-valued and integrable, and the constant \(Y\) is integrable and \(\mathcal{G}\)-measurable. There are only two events to check. For \(A=\varnothing\), both integrals are zero. For \(A=\Omega\), $$ \int_\Omega X\,d\mathbb{P}=\int_0^1x\,dx=\left[\frac{x^2}{2}\right]_0^1=\frac12, \qquad \int_\Omega Y\,d\mathbb{P}=\frac12\mathbb{P}(\Omega)=\frac12\cdot1=\frac12. $$ So \(Y\) meets the defining condition. Check: the value agrees with the average of \(x\) over the unit interval, and the theorem shows that any other integrable constant candidate must also equal \(1/2\) almost surely.
Worked Example: Different Versions on a Null Set
Let \(\Omega=[0,1]\), \(\mathcal{F}=\mathcal{B}_{[0,1]}\), let \(\mathbb{P}\) be Lebesgue probability measure, and take \(\mathcal{G}=\mathcal{F}\). For \(X=0\), compare \(Y=0\) with \(Z=\chi_{\{0\}}\).
Both candidates are real-valued, integrable, and \(\mathcal{G}\)-measurable. For every \(A\in\mathcal{G}\), \(\int_A Y\,d\mathbb{P}=0\), and $$ \int_A Z\,d\mathbb{P}=\int_A\chi_{\{0\}}\,d\mathbb{P} =\mathbb{P}(A\cap\{0\})=0, $$ because \(\mathbb{P}(\{0\})=0\). These are exactly the integrals of \(X\) over \(A\), so both candidates satisfy the condition. They differ at \(x=0\), but \(\mathbb{P}(\{0\})=0\), and hence \(Y=Z\) almost surely. Check: there is no pointwise uniqueness here; the conclusion of the theorem is precisely almost-sure uniqueness.
A Common Mistake: Checking Only the Whole Space
In the finite-partition example, \(\int_\Omega X\,d\mathbb{P}=2\). The constant candidate \(W=2\) also satisfies \(\int_\Omega W\,d\mathbb{P}=2\), since \(\mathbb{P}(\Omega)=1\). It is nevertheless not a conditional expectation given \(\mathcal{G}\). On the event \(C_1\), $$ \int_{C_1}W\,d\mathbb{P}=2\cdot\frac12=1, \qquad \int_{C_1}X\,d\mathbb{P}=\frac12. $$ The two values differ, so the required identity fails. Matching the unconditional expectation checks only the event \(\Omega\), not all events in \(\mathcal{G}\).
A second mistake is to read almost-sure uniqueness as pointwise uniqueness. The null-set example shows why that is too strong: two valid versions can differ at outcomes that together form a null set. Conversely, the uniqueness proof cannot be weakened to checking only a few selected events unless those checks establish the identity for every event in \(\mathcal{G}\). The event family is what detects the positive and negative sets of the difference.
Check Your Understanding
For each question, state which event-integral identities are being used and distinguish pointwise from almost-sure conclusions.
- On a four-point space with equal point probabilities and conditioning cells \(\{1,2\}\) and \(\{3,4\}\), let \(X\) take values \(4,2,0,2\) in that order. Find a candidate constant on each cell and verify its integrals on all four events in the conditioning sigma-algebra.
- If two integrable, \(\mathcal{G}\)-measurable candidates satisfy the event identities, why must the sets where their difference is at least \(1/n\) have probability zero for every positive integer \(n\)?
- Give an example of two candidates that satisfy the same event-integral identities but differ at one outcome. Explain why this does not contradict uniqueness.
- Why does equality of the two integrals over \(\Omega\) alone fail to establish the defining property when \(\mathcal{G}\) contains a proper nonempty event?
- In the Radon–Nikodym construction, identify the measure on \(\mathcal{G}\) whose derivative gives the conditional expectation, and state the identity its derivative satisfies.