The Pull-Out Property

Why the Pull-Out Property Matters

Conditional expectation averages a random variable while retaining the information recorded by a sigma-algebra. If a factor is already determined by that information, it does not need to be averaged again: it can be taken outside the conditional expectation.

This rule is useful in calculations involving products, but its hypotheses matter. A factor may be pulled out when it is measurable with respect to the sigma-algebra on which we condition; a factor that is merely measurable on the whole probability space need not be eligible. We first prove the bounded version, then give an extension that makes its integrability requirements explicit.

Setup and Definition

Let $(\Omega,\mathcal{F},P)$ be a probability space and let $\mathcal{G}\subseteq\mathcal{F}$ be a sigma-algebra. For an integrable real random variable $X$, a conditional expectation of $X$ given $\mathcal{G}$ is an integrable, $\mathcal{G}$-measurable random variable $Y$ such that, for every $A\in\mathcal{G}$,

\int_A Y\,dP=\int_A X\,dP.

It is unique up to almost-sure equality. We write $Y=E[X\mid\mathcal{G}]$. In the results below, equalities between conditional expectations are understood almost surely.

Theorem: If $X$ is integrable and $H$ is a bounded, real-valued, $\mathcal{G}$-measurable random variable, then $HX$ is integrable and $$ E[HX\mid\mathcal{G}]=H E[X\mid\mathcal{G}]. $$

Proof of the Bounded Pull-Out Property

Set $Y=E[X\mid\mathcal{G}]$. Since $H$ is bounded, $HX$ and $HY$ are integrable. Also, $HY$ is $\mathcal{G}$-measurable. It remains to verify the defining integral identity for $HY$.

Proof.

First take $H=\mathbf{1}_C$ for some $C\in\mathcal{G}$. For any $A\in\mathcal{G}$, the set $A\cap C$ belongs to $\mathcal{G}$, so the defining property of $Y$ gives

\int_A \mathbf{1}_C X\,dP =\int_{A\cap C}X\,dP =\int_{A\cap C}Y\,dP =\int_A\mathbf{1}_C Y\,dP.

By linearity of the integral, the same identity holds when $H$ is a $\mathcal{G}$-measurable simple function.

Now let $H$ be any bounded, $\mathcal{G}$-measurable real random variable. There are $\mathcal{G}$-measurable simple functions $H_n$ that converge uniformly to $H$. For every $A\in\mathcal{G}$, the simple-function case gives $\int_A H_nX\,dP=\int_A H_nY\,dP$. Moreover,

\int_\Omega |(H_n-H)X|\,dP \leq \|H_n-H\|_\infty\int_\Omega |X|\,dP \longrightarrow 0, $$ $$ \int_\Omega |(H_n-H)Y|\,dP \leq \|H_n-H\|_\infty\int_\Omega |Y|\,dP \longrightarrow 0.

Thus the integrals over $A$ converge on both sides, yielding $\int_A HX\,dP=\int_A HY\,dP$. Since this holds for every $A\in\mathcal{G}$, and $HY$ is integrable and $\mathcal{G}$-measurable, it is a conditional expectation of $HX$. Uniqueness proves the stated identity.

$\square$

What the Hypotheses Are Doing

The measurability condition says that $H$ is determined by the information in $\mathcal{G}$. In the proof, this allows sets of the form $A\cap C$ to remain in $\mathcal{G}$, so the defining integral identity for conditional expectation can be applied. Boundedness ensures that both products $HX$ and $HY$ are integrable.

In particular, $H$ need not be constant. It may vary across the information represented by $\mathcal{G}$. Nor does the identity say that an arbitrary factor can be pulled out: if a factor is not $\mathcal{G}$-measurable, the indicator-function argument does not apply.

Worked Example: A Factor on a Finite Partition

Let $\Omega=\{1,2,3,4\}$ with each point having probability $1/4$. Let $\mathcal{G}$ be generated by $C_1=\{1,2\}$ and $C_2=\{3,4\}$. Define $X$ by $X(1)=2$, $X(2)=6$, $X(3)=1$, and $X(4)=5$. Define $H=3$ on $C_1$ and $H=-2$ on $C_2$. The conditional expectation of $X$ on each partition cell is its average there:

E[X\mid\mathcal{G}](1)=E[X\mid\mathcal{G}](2) =\frac{2+6}{2}=4, \qquad E[X\mid\mathcal{G}](3)=E[X\mid\mathcal{G}](4) =\frac{1+5}{2}=3.

On $C_1$, the average of $HX$ is $(3\cdot2+3\cdot6)/2=12$, which equals $3\cdot4$. On $C_2$, it is $(-2\cdot1-2\cdot5)/2=-6$, which equals $(-2)\cdot3$. Therefore, cell by cell, $E[HX\mid\mathcal{G}]=H E[X\mid\mathcal{G}]$.

Products on a Continuous Probability Space

The same identity applies even when the information in $\mathcal{G}$ is described by a continuously distributed random variable. The defining test remains the integral identity over every event in $\mathcal{G}$.

Worked Example: A Bounded Function of the Conditioning Variable

Let $U$ and $V$ be independent uniform random variables on $[0,1]$, and let $\mathcal{G}=\sigma(U)$. Take $X=V$ and $H=1+U$. This factor is $\mathcal{G}$-measurable and bounded. For an event $A=\{U\in B\}$, where $B$ is a Borel subset of $[0,1]$, independence and integration give

E[\mathbf{1}_A V] =\int_B\int_0^1 v\,dv\,du =\frac12\int_B 1\,du.

Hence $E[V\mid\mathcal{G}]=1/2$. For the product,

E[\mathbf{1}_A(1+U)V] =\int_B\int_0^1(1+u)v\,dv\,du =\int_B\frac{1+u}{2}\,du.

The last expression is the integral over $A$ of $(1+U)/2$. Thus $E[(1+U)V\mid\mathcal{G}]=(1+U)/2$, exactly as the pull-out property predicts.

An Extension to Unbounded Factors

For an unbounded factor, integrability requires care. The following version gives a useful extension without assuming a general order or convergence property for conditional expectations. It explicitly requires that both products in the proposed identity be integrable.

Theorem: Let $X$ be integrable, let $Y=E[X\mid\mathcal{G}]$, and let $H$ be a finite-valued, $\mathcal{G}$-measurable random variable. If $HX$ and $HY$ are integrable, then $$ E[HX\mid\mathcal{G}]=HY. $$

Proof.

For $n\in\mathbb{N}$, define the clipped factor $H_n=\max(-n,\min(H,n))$. It is bounded and $\mathcal{G}$-measurable, so the bounded pull-out property gives $E[H_nX\mid\mathcal{G}]=H_nY$. In particular, for every $A\in\mathcal{G}$,

\int_A H_nX\,dP=\int_A H_nY\,dP.

Pointwise, $H_n\to H$, and $|H_n|\leq |H|$. Therefore $|H_nX|\leq|HX|$ and $|H_nY|\leq|HY|$. The assumed integrability of $HX$ and $HY$, together with the ordinary dominated convergence theorem, allows passage to the limit in both integrals. Consequently $\int_A HX\,dP=\int_A HY\,dP$ for every $A\in\mathcal{G}$. The random variable $HY$ is integrable by assumption and $\mathcal{G}$-measurable, so it satisfies the definition of $E[HX\mid\mathcal{G}]$. Uniqueness gives the result.

$\square$

Worked Example: An Unbounded Factor with Integrable Products

Let $U,V$ again be independent uniform random variables on $[0,1]$, and set $\mathcal{G}=\sigma(U)$, $X=V$. Define $H=U^{-1/2}$ when $U>0$, and set $H=0$ when $U=0$. Then $H$ is finite almost surely and $\mathcal{G}$-measurable. Since $E[V\mid\mathcal{G}]=1/2$, both products are integrable:

E[|HX|] =\int_0^1\int_0^1 u^{-1/2}v\,dv\,du =\frac12\cdot2=1, \qquad E[|H E[X\mid\mathcal{G}]|] =\frac12\int_0^1u^{-1/2}\,du=1.

The unbounded-factor theorem therefore applies and gives $E[HV\mid\mathcal{G}]=H/2$. To check the result directly, for $A=\{U\in B\}$,

E[\mathbf{1}_A HV] =\int_B\int_0^1u^{-1/2}v\,dv\,du =\frac12\int_Bu^{-1/2}\,du =E[\mathbf{1}_A H/2].

The improper integral is finite even when $B=[0,1]$, since $\int_0^1u^{-1/2}\,du=2$. The value assigned at $U=0$ does not affect the calculation because that event has probability zero.

A Practical Check Before Pulling Out a Factor

Before using the identity, check three things: the factor is measurable with respect to the conditioning sigma-algebra; the product inside the conditional expectation is integrable; and, for an unbounded factor, the proposed outside product is also integrable. For bounded factors and integrable $X$, the last two integrability requirements follow immediately from boundedness. In the unbounded case, the extension proved above states the conditions directly rather than silently assuming them.

A common mistake is to treat a factor as pull-out eligible merely because it is known as a random variable on the original probability space. The relevant question is more specific: is the factor measurable with respect to the information we are conditioning on? If not, the bounded pull-out theorem does not apply.

Check Your Understanding

Use the hypotheses of the results above to answer these questions.

Why does $\mathcal{G}$-measurability of $H$ allow the indicator-function proof to use the defining property of conditional expectation?
Which integrability requirements are automatic when $H$ is bounded and $X$ is integrable?
In the unbounded-factor theorem, why are integrability of $HX$ and of $H E[X\mid\mathcal{G}]$ both specified?
For independent uniform $U,V$ and $\mathcal{G}=\sigma(U)$, what is $E[(2+U)V\mid\mathcal{G}]$?
What condition on a factor fails if it depends on information not contained in $\mathcal{G}$?

What You'll Learn