Conditional Expectation and Order
Conditional expectation averages an integrable random variable relative to the information in a sub-sigma-algebra. Its defining integral identity might not appear to impose any pointwise sign or order on the resulting function. Nevertheless, it does impose both: nonnegative random variables have nonnegative conditional expectations almost surely, and almost-sure inequalities are preserved. The qualification “almost surely” matters because conditional expectations are determined only up to changes on null sets.
Throughout, let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, not assumed complete, and let \(\mathcal{G}\subseteq\mathcal{F}\) be a sub-sigma-algebra. Random variables in this tutorial are real-valued and integrable, meaning \(\mathbb{E}[|X|]<\infty\). An assertion holds almost surely if the set where it fails has probability zero. We use the earlier characterization of conditional expectation: for integrable \(X\), \(\mathbb{E}[X\mid\mathcal{G}]\) is a \(\mathcal{G}\)-measurable integrable random variable \(Z\) satisfying the defining identity below. Its existence follows from the Radon–Nikodym construction, and any two such random variables are equal almost surely.
The requirement is an identity on every event in \(\mathcal{G}\), not a pointwise formula. In particular, when \(\mathcal{G}\) is generated by a finite partition, a conditional expectation is constant on each positive-probability atom, and that constant is the probability-weighted average of \(X\) on the atom. Values on zero-probability atoms are not fixed by the identity.
Computing on a Finite Partition
Worked Example: A Nonnegative Conditional Expectation
Let \(\Omega=\{1,2,3,4\}\), assign probability \(1/4\) to each outcome, and let \(\mathcal{G}\) be generated by \(C_1=\{1,2\}\) and \(C_2=\{3,4\}\). Define \(X(1)=0,\ X(2)=4,\ X(3)=2,\ X(4)=6\). Compute \(\mathbb{E}[X\mid\mathcal{G}]\) and check the defining identity.
A \(\mathcal{G}\)-measurable function is constant on each of \(C_1,C_2\). On \(C_1\), the required constant is the average of \(0\) and \(4\), so it is \((0+4)/2=2\). On \(C_2\), it is the average of \(2\) and \(6\), so it is \((2+6)/2=4\). Thus take \(Z=2\chi_{C_1}+4\chi_{C_2}\). It is nonnegative.
There are four events in \(\mathcal{G}\). On \(\varnothing\), both integrals are \(0\). On \(C_1\), \(\int_{C_1}X\,d\mathbb{P}=(0+4)/4=1\), while \(\int_{C_1}Z\,d\mathbb{P}=2(2/4)=1\). On \(C_2\), \(\int_{C_2}X\,d\mathbb{P}=(2+6)/4=2\), while \(\int_{C_2}Z\,d\mathbb{P}=4(2/4)=2\). On \(\Omega\), the respective integrals are \(0+1+2=3\) and \(1+2=3\). These checks verify the identity on every event in \(\mathcal{G}\). As a further check, the two atom averages, \(2\) and \(4\), are both nonnegative, as expected from the nonnegative values of \(X\).
Positivity and Preservation of Order
Proof. Write \(Z=\mathbb{E}[X\mid\mathcal{G}]\), choosing a \(\mathcal{G}\)-measurable integrable representative. For positivity, let \(N=\{Z<0\}\), which belongs to \(\mathcal{G}\). The defining identity gives \(\int_N Z\,d\mathbb{P}=\int_N X\,d\mathbb{P}\geq0\), since \(X\geq0\) almost surely. Suppose \(\mathbb{P}(N)>0\). The sets \(N_n=\{Z\leq-1/n\}\), \(n\geq1\), have union \(N\): every strictly negative real number is at most \(-1/n\) for some \(n\). Countable subadditivity implies that some \(N_n\) has positive probability. Since \(Z<0\) on \(N\) and \(Z\leq-1/n\) on \(N_n\), monotonicity of the integral gives \(\int_N Z\,d\mathbb{P}\leq-\mathbb{P}(N_n)/n<0\), a contradiction. Therefore \(\mathbb{P}(N)=0\), proving positivity almost surely.
For order preservation, choose representatives \(U=\mathbb{E}[X\mid\mathcal{G}]\) and \(V=\mathbb{E}[Y\mid\mathcal{G}]\), and set \(A=\{U>V\}\in\mathcal{G}\). The defining identities and monotonicity of the integral give $$\int_A U\,d\mathbb{P}=\int_A X\,d\mathbb{P}\leq\int_A Y\,d\mathbb{P}=\int_A V\,d\mathbb{P}.$$ The inequality holds because \(X\leq Y\) almost surely. If \(\mathbb{P}(A)>0\), then \(A\) is the union of the sets \(A_n=\{U-V\geq1/n\}\), \(n\geq1\), so some \(A_n\) has positive probability. The integrable function \(U-V\) is positive on \(A\) and at least \(1/n\) on \(A_n\); hence \(\int_A(U-V)\,d\mathbb{P}\geq\mathbb{P}(A_n)/n>0\). By linearity of the integral this contradicts the displayed inequality. Thus \(\mathbb{P}(A)=0\), proving \(U\leq V\) almost surely. \(\square\)
The proof uses only the defining identities on events belonging to \(\mathcal{G}\), along with ordinary monotonicity and linearity of integration. In particular, it does not require the probability space or \(\mathcal{G}\) to be complete. The conclusion concerns almost-sure order, not necessarily pointwise order of every chosen representative.
Worked Example: Order with a Null Atom
Let \(\Omega=\{a,b,c\}\), with \(\mathbb{P}(\{a\})=\mathbb{P}(\{b\})=1/2\) and \(\mathbb{P}(\{c\})=0\). Let \(\mathcal{G}\) consist of all unions of \(C=\{a,b\}\) and \(N=\{c\}\). Set \(X(a)=0,\ X(b)=2,\ X(c)=100\), and \(Y(a)=1,\ Y(b)=3,\ Y(c)=-50\). Then \(X\leq Y\) on \(C\), so \(X\leq Y\) almost surely, even though the inequality fails at \(c\).
On \(C\), the average of \(X\) is \((0+2)/2=1\), and the average of \(Y\) is \((1+3)/2=2\). Define \(U\) to equal \(1\) on \(C\) and \(9\) on \(N\); define \(V\) to equal \(2\) on \(C\) and \(5\) on \(N\). These are \(\mathcal{G}\)-measurable integrable versions of the respective conditional expectations. To check this, \(\mathcal{G}\) has the four events \(\varnothing,C,N,\Omega\). On \(C\), the integrals of \(X,U\) are both \(1\), and those of \(Y,V\) are both \(2\). On \(N\), all four integrals are \(0\), because \(\mathbb{P}(N)=0\); on \(\Omega\), the integrals match the ones on \(C\). They also match on \(\varnothing\). Thus the defining identities hold for every event in \(\mathcal{G}\).
On \(N\), these chosen versions satisfy \(U(c)=9>5=V(c)\), so their pointwise order fails there. This does not contradict the theorem: the failure occurs only on a null set, and \(U\leq V\) almost surely. The example also shows why the values on a zero-probability atom cannot be recovered from the defining integral identities.
Worked Example: Positivity and a Changed Version
Take \(\Omega=\{r,s\}\), with \(\mathbb{P}(\{r\})=1\), \(\mathbb{P}(\{s\})=0\), and \(\mathcal{F}=\mathcal{G}\) equal to all subsets of \(\Omega\). Let \(X(r)=3\) and \(X(s)=0\), and define \(Z(r)=3,\ Z(s)=-4\). The random variable \(X\) is nonnegative everywhere, and \(Z\) is a \(\mathcal{G}\)-measurable integrable version of \(\mathbb{E}[X\mid\mathcal{G}]\).
Indeed, for the four events \(\varnothing,\{r\},\{s\},\Omega\), the integrals of \(X\) and \(Z\) are respectively \(0,3,0,3\). Thus the identities hold on all of \(\mathcal{G}\). Yet \(Z(s)=-4\). The theorem says \(Z\geq0\) almost surely, which is true because the only point where \(Z<0\) has probability zero. As an independent check, the representative \(Z_0=X\) is also a conditional expectation, and \(Z=Z_0\) except on \(\{s\}\), a null set.
What the Almost-Sure Qualification Does
A frequent mistake is to turn the theorem into a pointwise claim about a particular version. In the last example, the incorrect calculation is: “\(X\geq0\), so \(Z(s)\geq0\); therefore \(Z(s)=-4\) cannot be a conditional expectation.” The correction is that positivity has been proved almost surely, not at every outcome. The defining identity tests integrals over events, and changing a value at a zero-probability outcome changes none of those integrals.
The same distinction applies to order. If \(X\leq Y\) only almost surely, their values on the exceptional null set need not be ordered. Even when representatives of the conditional expectations are chosen, the theorem guarantees their order only almost surely. One may choose versions that satisfy the order everywhere by changing them on an appropriate \(\mathcal{G}\)-measurable null set when such a modification is available, but no such pointwise choice is needed for the theorem. The order conclusion itself is an almost-sure statement.
These results are useful because order information survives conditioning without requiring a formula for the conditional expectation. For example, if \(0\leq X\leq c\) almost surely for a constant \(c\), then applying the theorem to \(0\leq X\) and \(X\leq c\) gives \(0\leq\mathbb{E}[X\mid\mathcal{G}]\leq c\) almost surely. This deduction uses the order theorem twice; it does not assert that any fixed representative obeys the bounds at every point.
Check Your Understanding
In each question, distinguish identities that hold everywhere as functions from conclusions that hold only almost surely.
- On a finite probability space with a two-atom conditioning partition, explain why the conditional expectation is constant on each positive-probability atom.
- If \(X\geq0\) almost surely, what event is used in the positivity proof, and why does its positive probability lead to a contradiction?
- If \(X\leq Y\) almost surely, explain why the event \(\{\mathbb{E}[X\mid\mathcal{G}]>\mathbb{E}[Y\mid\mathcal{G}]\}\) belongs to \(\mathcal{G}\).
- Can two versions of the same conditional expectation have different values at a point of positive probability? Justify your answer using almost-sure uniqueness.
- In the null-atom example, why do the defining integral identities fail to restrict the chosen conditional-expectation values on \(N\)?