The Pull-Out Property for Known Quantities

Why a Known Factor Can Be Pulled Out

A conditional expectation averages a random variable while holding the conditioning information fixed. If a factor is already determined by that information, it has the same value throughout each conditional average. It can therefore multiply the average rather than remain inside it. The phrase “known given the conditioning information” has a precise meaning: the factor must be measurable with respect to the information on which we condition.

This principle is called the pull-out property. Its familiar discrete form says that if we condition on a discrete random variable \(Z\), then a function of \(Z\) can be taken outside the conditional expectation. The general result applies to any sigma-algebra and, with an integrability condition, even to an unbounded factor.

The defining averaging condition is the main tool in the proof. Informally, it says that averaging \(X\) over any event described by the conditioning information gives the same result as averaging its conditional expectation over that event. The pull-out property shows how this characterization behaves when we multiply by a factor already described by \(\mathcal G\).

The Pull-Out Theorem

Here “finite-valued” means that \(Y\) is real-valued almost surely; it does not mean that it takes only finitely many distinct values. The integrability assumptions ensure that both sides define integrable random variables. To prove the result carefully, we first establish the averaging identity for a bounded \(\mathcal G\)-measurable multiplier.

Proof of the lemma. First take \(H\) to be a simple random variable, \(H=\sum_{j=1}^m c_j 1_{A_j}\), where \(A_j\in\mathcal G\) and the \(c_j\) are real constants. By the defining averaging condition and linearity,

For bounded measurable \(H\), choose simple \(\mathcal G\)-measurable \(H_k\) such that \(H_k\to H\) pointwise and \(|H_k|\leq C\) for a finite constant \(C\). The dominated convergence theorem applies on the left, since \(|H_k U|\leq C|U|\). It applies on the right because \(E[U\mid\mathcal G]\) is integrable and \(|H_k E[U\mid\mathcal G]|\leq C|E[U\mid\mathcal G]|\). Taking limits proves the lemma. \(\square\)

We now prove the theorem, including the case where \(Y\) is unbounded. Write \(M=E[X\mid\mathcal G]\), and define \(Y_n=\max(-n,\min(Y,n))\). Each \(Y_n\) is bounded and \(\mathcal G\)-measurable. The lemma, applied to \(U=|X|\) and \(H=\min(|Y|,n)\), gives

Both sides increase as \(n\) increases. By the monotone convergence theorem, their limits are equal, so

The inequality \(|M|\leq E[|X|\mid\mathcal G]\) holds almost surely by the standard conditional-expectation bound. Consequently, \(E[|YM|]\leq E[|Y|E[|X|\mid\mathcal G]]<\infty\). Thus \(YM\) is integrable, as is \(YX\) by assumption.

For any \(A\in\mathcal G\), the random variable \(1_A Y_n\) is bounded and \(\mathcal G\)-measurable. Applying the lemma with \(U=X\) and \(H=1_A Y_n\) yields \(E[1_A Y_n X]=E[1_A Y_n M]\). As \(n\) tends to infinity, \(Y_n\to Y\) pointwise and \(|Y_n|\leq |Y|\). The dominated convergence theorem applies to both sides: the dominating variables are \(1_A|YX|\) and \(1_A|YM|\), respectively, both integrable. Hence

Since \(YM\) is \(\mathcal G\)-measurable and integrable, this is exactly the defining characterization of \(E[YX\mid\mathcal G]\). Therefore \(E[YX\mid\mathcal G]=YM\) almost surely. \(\square\)

Discrete Conditioning: A Factor That Is a Function of the Conditioning Variable

Suppose \(Z\) is discrete and we condition on \(Z\). A function \(g(Z)\) is determined by the observed value of \(Z\), so it is measurable with respect to the information generated by \(Z\). When \(X\) and \(g(Z)X\) are integrable, the theorem gives

At a particular value \(z\) with \(P(Z=z)>0\), this is the ordinary conditional-mean identity \(E[g(z)X\mid Z=z]=g(z)E[X\mid Z=z]\): within that conditional average, \(g(z)\) is a constant. Values \(z\) with probability zero do not determine a conditional expectation and need not be assigned a particular conditional mean.

Finite Partitions: The Same Identity on Each Atom

A finite partition is a collection of disjoint events \(A_1,\ldots,A_m\) whose union is the entire sample space. Conditioning on the partition means knowing which event \(A_j\) occurred. A random variable \(Y\) is measurable with respect to this information when it is constant on each partition event, apart from possible null sets. On each event, pulling out \(Y\) is simply pulling out a numerical constant.

The finite-partition calculation extends to any \(\mathcal G\)-measurable factor, not just one that is constant on a finite number of events. In the general setting, measurability plays the role of “constant on each conditioning case.” The proof above replaces a finite list of cases with bounded approximations and the defining averaging property.

Unbounded Factors and the Integrability Check

Boundedness of \(Y\) is a convenient sufficient condition, but it is not necessary. For an unbounded factor, the theorem requires \(XY\) to be integrable. The proof also establishes that \(Y E[X\mid\mathcal G]\) is integrable under these assumptions. Checking this condition matters: an identity involving conditional expectations of nonintegrable variables is not justified by the pull-out property.

Common Mistake: Pulling Out a Factor That Is Not Known

A factor cannot be pulled out merely because it is a random variable in the same model, or because its unconditional value is known. It must be determined by the conditioning information. For example, if \(X\) and \(Y\) are independent Bernoulli random variables with success probability \(1/2\), and we condition on \(X\), then \(Y\) is not determined by \(X\). In fact,

These are not equal almost surely: when \(X=1\) and \(Y=0\), which has probability \(1/4\), the first is \(1/2\) and the second is \(0\). The incorrect reasoning is “\(X\) is known given \(X\), so it can be replaced by any factor in the product.” The correction is to check the measurability of the factor being pulled out: here \(X\) is known given \(X\), but \(Y\) is not.