When a Factor Can Be Taken Outside
Conditional expectation averages over the information not recorded by a conditioning sigma-algebra. If a factor \(H\) is already measurable with respect to that sigma-algebra, it is fixed by that averaging and can be taken outside the conditional expectation. The statement has important qualifications: the integrand must be integrable, the factor must be measurable with respect to the conditioning sigma-algebra, and for an unbounded factor the product must be integrable.
Throughout, \((\Omega,\mathcal{F},\mathbb{P})\) is a probability space, not assumed complete, and \(\mathcal{G}\) is a sub-sigma-algebra of \(\mathcal{F}\). All random variables in this tutorial are real-valued and measurable; integrability means finite expectation of the absolute value. Equalities between conditional expectations are understood \(\mathbb{P}\)-almost surely. For \(Y\in L^1(\mathbb{P})\), a conditional expectation \(Z=\mathbb{E}[Y\mid\mathcal{G}]\) is an integrable, \(\mathcal{G}\)-measurable random variable satisfying \(\int_A Z\,d\mathbb{P}=\int_A Y\,d\mathbb{P}\) for every \(A\in\mathcal{G}\); it is unique almost surely.
We use the characterization and order properties of conditional expectation established earlier: it is linear, preserves nonnegativity, and \(\mathbb{E}[|Y|\mid\mathcal{G}]\) is defined for integrable \(Y\). In particular, if \(Z=\mathbb{E}[Y\mid\mathcal{G}]\), then \(|Z|\leq\mathbb{E}[|Y|\mid\mathcal{G}]\) almost surely. This follows by applying order preservation to \(-|Y|\leq Y\leq |Y|\).
Proof. First suppose \(H\) is simple, written as \(H=\sum_{j=1}^k c_j\chi_{B_j}\), where the \(B_j\in\mathcal{G}\) are disjoint and cover \(\Omega\). For every \(A\in\mathcal{G}\), the defining property of \(Z=\mathbb{E}[Y\mid\mathcal{G}]\) gives $$\int_A HY\,d\mathbb{P}=\sum_{j=1}^k c_j\int_{A\cap B_j}Y\,d\mathbb{P} =\sum_{j=1}^k c_j\int_{A\cap B_j}Z\,d\mathbb{P} =\int_A HZ\,d\mathbb{P}.$$ Also \(HY\) and \(HZ\) are integrable because \(H\) is bounded and \(Y,Z\in L^1\). Thus the characterization of conditional expectation proves the identity for simple \(H\).
For general \(H\), first address the distinction between bounded everywhere and bounded almost surely. Choose \(C<\infty\) with \(\mathbb{P}(|H|>C)=0\), let \(N=\{|H|>C\}\in\mathcal{G}\), and define \(H'=H\) on \(N^c\) and \(H'=0\) on \(N\). Then \(H'\) is \(\mathcal{G}\)-measurable, bounded everywhere, and equal to \(H\) almost surely. The products \(H'Y=HY\) and \(H'Z=HZ\) therefore agree almost surely.
Approximate \(H'\) by simple, \(\mathcal{G}\)-measurable functions \(H_n\) whose values lie in a bounded interval and satisfy \(|H_n-H'|\leq 1/n\) everywhere. Such functions are obtained by dividing the interval \([-C,C]\) into finitely many measurable subintervals of length at most \(1/n\), and assigning one value to each inverse image. The simple-factor case gives \(\int_A H_nY\,d\mathbb{P}=\int_A H_nZ\,d\mathbb{P}\) for every \(A\in\mathcal{G}\). Moreover, $$\left|\int_A(H_n-H')Y\,d\mathbb{P}\right|\leq\frac1n\mathbb{E}|Y|,\qquad \left|\int_A(H_n-H')Z\,d\mathbb{P}\right|\leq\frac1n\mathbb{E}|Z|\leq\frac1n\mathbb{E}|Y|.$$ The last inequality follows from \(|Z|\leq\mathbb{E}[|Y|\mid\mathcal{G}]\) and the defining property with \(A=\Omega\). Letting \(n\to\infty\) proves \(\int_A H'Y\,d\mathbb{P}=\int_A H'Z\,d\mathbb{P}\) for every \(A\in\mathcal{G}\). The characterization now yields \(\mathbb{E}[H'Y\mid\mathcal{G}]=H'Z\) almost surely. Replacing \(H'\) by its almost-sure equal version \(H\) proves the result. Finally, \(HY\) is integrable since \(\mathbb{E}|HY|\leq C\mathbb{E}|Y|\). \(\square\)
Worked Example: A Factor on a Finite Partition
Let \(\Omega=\{a,b,c,d\}\), give each point probability \(1/4\), and let \(\mathcal{G}\) be generated by the partition \(\{\{a,b\},\{c,d\}\}\). Set \(Y(a)=2,\ Y(b)=6,\ Y(c)=-1,\ Y(d)=3\), and let \(H=5\) on \(\{a,b\}\) and \(H=-2\) on \(\{c,d\}\). The factor \(H\) is bounded and \(\mathcal{G}\)-measurable.
On the first partition cell, the conditional average of \(Y\) is \((2+6)/2=4\). On the second it is \((-1+3)/2=1\). Thus \(Z=\mathbb{E}[Y\mid\mathcal{G}]\) takes values \(4,4,1,1\), in the order \(a,b,c,d\), and \(HZ\) takes values \(20,20,-2,-2\).
Directly, \(HY\) takes values \(10,30,2,-6\). Its conditional averages on the two cells are \((10+30)/2=20\) and \((2-6)/2=-2\), as predicted. Check: \(\mathbb{E}[HY]=(10+30+2-6)/4=36/4=9\), while \(\mathbb{E}[HZ]=(20+20-2-2)/4=36/4=9\). The cellwise conditional averages, not just the overall expectations, agree.
Unbounded Factors: The Integrability Condition
Boundedness is a convenient sufficient condition, not the only one. If \(H\) is unbounded, the product \(HY\) must at least be integrable for the ordinary conditional expectation \(\mathbb{E}[HY\mid\mathcal{G}]\) to be defined. Under that condition the pull-out identity still holds, and its right-hand side is automatically integrable as well.
Proof. Put \(Z=\mathbb{E}[Y\mid\mathcal{G}]\). For each positive integer \(n\), the bounded, \(\mathcal{G}\)-measurable factor \(|H|\wedge n\) satisfies the bounded pull-out property applied to \(|Y|\). Consequently, $$\mathbb{E}\big[(|H|\wedge n)\mathbb{E}[|Y|\mid\mathcal{G}]\big] =\mathbb{E}\big[(|H|\wedge n)|Y|\big]\leq\mathbb{E}[|HY|].$$ Since \(|Z|\leq\mathbb{E}[|Y|\mid\mathcal{G}]\) almost surely, the same upper bound holds for \(\mathbb{E}[(|H|\wedge n)|Z|]\). The nonnegative random variables \((|H|\wedge n)|Z|\) increase to \(|H||Z|\). By the Monotone Convergence Theorem, $$\mathbb{E}[|H||Z|]=\lim_{n\to\infty}\mathbb{E}[(|H|\wedge n)|Z|]\leq\mathbb{E}[|HY|]<\infty.$$ Thus \(HZ\) is integrable.
Now set \(H_n=\max(-n,\min(H,n))\). These are bounded and \(\mathcal{G}\)-measurable, \(H_n\to H\) pointwise, and \(|H_n|\leq|H|\). By the bounded result, \(\mathbb{E}[H_nY\mid\mathcal{G}]=H_nZ\) almost surely for each \(n\). Dominated Convergence applies to \(H_nY\to HY\), since \(|H_nY-HY|\leq2|HY|\), and to \(H_nZ\to HZ\), since \(|H_nZ-HZ|\leq2|HZ|\). Hence both convergences hold in \(L^1\). Conditional expectation is an \(L^1\) contraction: \(\mathbb{E}|\mathbb{E}[U\mid\mathcal{G}]|\leq\mathbb{E}|U|\), by the conditional order inequality and integration. Therefore \(\mathbb{E}[H_nY\mid\mathcal{G}]\to\mathbb{E}[HY\mid\mathcal{G}]\) in \(L^1\), while \(H_nZ\to HZ\) in \(L^1\). Since the two terms are equal almost surely for every \(n\), their \(L^1\) limits are equal almost surely. This proves the claim. \(\square\)
Worked Example: An Unbounded Factor on a Countable Space
Let \(\Omega=\mathbb{N}=\{1,2,\ldots\}\), with \(\mathbb{P}(\{n\})=2^{-n}\). These masses sum to \(1\). Let \(\mathcal{G}\) be generated by the pairs \(B_k=\{2k-1,2k\}\), \(k\geq1\). Define \(H(n)=k\) on \(B_k\), and set \(Y(2k-1)=1,\ Y(2k)=3\). Then \(H\) is finite-valued, \(\mathcal{G}\)-measurable, and unbounded.
The two point masses in \(B_k\) are \(2^{-(2k-1)}=2\cdot4^{-k}\) and \(2^{-2k}=4^{-k}\). Their total is \(3\cdot4^{-k}\), so the conditional average of \(Y\) on this pair is $$(2\cdot4^{-k}\cdot1+4^{-k}\cdot3)/(3\cdot4^{-k})=5/3.$$ Thus \(Z=\mathbb{E}[Y\mid\mathcal{G}]\) equals \(5/3\) on every pair.
The product is integrable: summing its nonnegative contributions pair by pair gives $$\mathbb{E}|HY|=\sum_{k=1}^{\infty}k\big(2\cdot4^{-k}\cdot1+4^{-k}\cdot3\big) =5\sum_{k=1}^{\infty}\frac{k}{4^k} =5\cdot\frac{4}{9}=\frac{20}{9}.$$ Here \(\sum_{k\geq1}kr^k=r/(1-r)^2\) for \(0\leq r<1\), obtained by differentiating the geometric series; at \(r=1/4\) it is \((1/4)/(3/4)^2=4/9\). The right-hand side \(HZ\) has expectation \(\sum_{k\geq1}(3\cdot4^{-k})k(5/3)=5\sum_{k\geq1}k/4^k=20/9\). More strongly, on each pair the conditional average of \(HY\) is \((2\cdot4^{-k}k+4^{-k}3k)/(3\cdot4^{-k})=5k/3\), exactly \(HZ\). These pairwise averages verify the identity directly. Check: both sides have finite expectation \(20/9\), and their values on every conditioning cell agree.
Why the Hypotheses Matter
Worked Example: A Bounded-A.S. Factor with an Unbounded Range
On \([0,1]\) with Lebesgue probability measure and \(\mathcal{G}=\mathcal{F}\) the Lebesgue sigma-algebra, define \(H(1/j)=j\) for integers \(j\geq2\), and \(H(x)=0\) at all other points. This is measurable and has an unbounded range, but it is zero almost surely: the exceptional set \(\{1/j:j\geq2\}\) is countable and has measure zero. Take \(Y(x)=1+x\), which is integrable.
Because \(\mathcal{G}=\mathcal{F}\), \(\mathbb{E}[Y\mid\mathcal{G}]=Y\) almost surely. Both \(HY\) and \(H\mathbb{E}[Y\mid\mathcal{G}]=HY\) vanish off the countable exceptional set, so each has integral \(0\). The identity holds almost surely. To approximate a bounded-almost-sure factor in the bounded proof, one must first replace it by a bounded-everywhere version, as in the proof above; a finite-valued simple function cannot approximate this unbounded range uniformly everywhere.
Measurability is essential. On a two-point space with equal probabilities, take \(\mathcal{G}=\{\varnothing,\Omega\}\), \(Y(a)=1,\ Y(b)=-1\), and \(H=Y\). Here \(\mathbb{E}[Y\mid\mathcal{G}]=0\), so \(H\mathbb{E}[Y\mid\mathcal{G}]=0\). But \(HY=1\) everywhere, and \(\mathbb{E}[HY\mid\mathcal{G}]=1\). The factor \(H\) is not \(\mathcal{G}\)-measurable, so the theorem does not apply.
Product integrability is essential for the unbounded version. On \((0,1)\) with Lebesgue probability measure and full conditioning sigma-algebra, let \(Y=1\) and \(H(x)=1/x\). Then \(H\) is finite and measurable, but \(\int_0^1|HY|\,dx=\int_0^1 dx/x=\infty\). The ordinary integrable conditional expectation of \(HY\) is not defined, so the stated unbounded-factor theorem cannot be invoked.
The practical test is to identify the conditioning sigma-algebra before manipulating a conditional expectation. A bounded factor measurable with respect to it may always be pulled out when the integrand is integrable. An unbounded factor may also be pulled out if its product with the integrand is integrable. Neither condition permits pulling out a factor that is not measurable with respect to the conditioning sigma-algebra.
Check Your Understanding
For each question, state which hypotheses apply and interpret equalities as almost-sure equalities.
- If \(H\) is \(\mathcal{G}\)-measurable and \(|H|\leq 4\) almost surely, while \(Y\in L^1(\mathbb{P})\), why is \(HY\) integrable?
- In the finite-partition example, compute \(\mathbb{E}[HY\mid\mathcal{G}]\) on each of the two partition cells.
- For an unbounded \(\mathcal{G}\)-measurable \(H\), which integrability condition in the extended pull-out theorem ensures that both sides are well-defined?
- Why must the bounded-almost-sure proof replace \(H\) by a bounded-everywhere version before using uniform simple-function approximations?
- Give the values of \(\mathbb{E}[HY\mid\mathcal{G}]\) and \(H\mathbb{E}[Y\mid\mathcal{G}]\) in the two-point example where \(H=Y\) and \(\mathcal{G}\) is trivial. Which hypothesis fails?