Why an L1 Contraction Matters
Conditional expectation replaces a random variable by an average that retains only the information represented by a sigma-algebra. Averaging can smooth out variation, but it cannot increase the average absolute size of a random variable. The precise statement is an inequality in the \(L^1\) norm.
This inequality is useful whenever a random variable is approximated by another one. If the approximation error is small on average, then conditioning both variables on the same information cannot make their average error larger. It also gives a way to compare conditional averages at different levels of information.
The \(L^1\) Norm and Conditional Expectation
A real-valued random variable \(X\) is integrable if \(\mathbb{E}[|X|]<\infty\). Its \(L^1\) norm is
As usual, random variables that are equal almost surely represent the same element of \(L^1\), and the norm is understood on these equivalence classes. For a sigma-algebra \(\mathcal{G}\), the conditional expectation \(\mathbb{E}[X\mid\mathcal{G}]\) is an integrable, \(\mathcal{G}\)-measurable random variable whose integral over each event in \(\mathcal{G}\) agrees with the integral of \(X\): for every \(A\in\mathcal{G}\),
Conditional expectations are determined only up to almost-sure equality. All norm statements below are therefore independent of the chosen version.
The key point behind this inequality is that the absolute value of a conditional average is bounded by the conditional average of the absolute value. We establish that fact first.
The \(L^1\) Contraction Theorem
The theorem says that conditional expectation, viewed as a map on \(L^1\), has operator norm at most one. In particular, it is continuous in the \(L^1\) norm. The same inequality applies to differences: linearity of conditional expectation gives the following consequence.
Worked Examples
Worked Example: Averaging on a Finite Partition
Let \(\Omega=\{1,2,3,4\}\), with probability \(1/4\) at each point. Let \(\mathcal{G}\) be the sigma-algebra generated by the partition \(\{\{1,2\},\{3,4\}\}\), and let \(X\) take the values \(-2,2,1,5\) at the four points.
On each cell of the partition, the conditional expectation is the average of \(X\) on that cell. Thus \(\mathbb{E}[X\mid\mathcal{G}]\) takes value \((-2+2)/2=0\) on \(\{1,2\}\), and value \((1+5)/2=3\) on \(\{3,4\}\). Its norm is
Meanwhile, \(\|X\|_1=(2+2+1+5)/4=5/2\). Hence \(3/2\leq 5/2\), as the contraction theorem requires. The first cell illustrates how averaging can reduce absolute size: opposite signs cancel in the conditional average, although both values contribute positively to the average of \(|X|\).
Worked Example: Controlling an Approximation Error
Use the same four-point probability space and the same sigma-algebra \(\mathcal{G}\). Let \(X=(2,0,4,2)\) and \(Y=(0,0,2,2)\), where each tuple lists the values at points \(1,2,3,4\), respectively.
On the two partition cells, the conditional averages are \(\mathbb{E}[X\mid\mathcal{G}]=(1,1,3,3)\) and \(\mathbb{E}[Y\mid\mathcal{G}]=(0,0,2,2)\). Their difference has absolute value \(1\) at every point, so
The original difference is \(X-Y=(2,0,2,0)\), and therefore \(\|X-Y\|_1=(2+0+2+0)/4=1\). The inequality holds with equality. In general, the corollary ensures that replacing \(X\) by an approximation \(Y\) with small \(L^1\) error also approximates their conditional expectations in \(L^1\), with no larger error.
Comparing Conditional Averages
The difference inequality compares conditional averages formed using the same sigma-algebra. There is also a useful comparison when one sigma-algebra contains another. If \(\mathcal{G}\subseteq\mathcal{H}\), then \(\mathcal{G}\) represents less information than \(\mathcal{H}\). The tower property says that conditioning an \(\mathcal{H}\)-conditional expectation again on \(\mathcal{G}\) gives the \(\mathcal{G}\)-conditional expectation.
Worked Example: Comparing Two Levels of Information
Let \(\Omega=\{1,2,3,4\}\) again have equal point probabilities. Let \(\mathcal{H}\) be generated by \(\{\{1,2\},\{3,4\}\}\), and let \(\mathcal{G}=\{\varnothing,\Omega\}\), so \(\mathcal{G}\subseteq\mathcal{H}\). Take \(X=(-1,3,2,6)\).
The \(\mathcal{G}\)-conditional expectation is the constant overall mean, \(\mathbb{E}[X\mid\mathcal{G}]=(-1+3+2+6)/4=5/2\). The \(\mathcal{H}\)-conditional expectation is \(1\) on \(\{1,2\}\) and \(4\) on \(\{3,4\}\), since the cell averages are \((-1+3)/2=1\) and \((2+6)/2=4\). Thus
The residual from the finer conditional average is \((-2,2,-2,2)\), so its \(L^1\) norm is \(2\). The comparison theorem gives \(3/2\leq 2\), as verified directly.
For sigma-algebras that are not nested, the same one-sided comparison need not apply. A general bound follows from the triangle inequality: for integrable \(X\) and sigma-algebras \(\mathcal{G}\) and \(\mathcal{H}\),
Indeed, insert \(X\) between the two conditional expectations and apply the triangle inequality. Thus, two conditional averages are close whenever each is close to the original variable in \(L^1\). The nested-sigma-algebra theorem is stronger because it controls their difference using only the residual from the finer conditioning.
Expectation Errors and a Common Pitfall
Taking \(\mathcal{G}=\{\varnothing,\Omega\}\) gives \(\mathbb{E}[X\mid\mathcal{G}]=\mathbb{E}[X]\), a constant. The difference inequality then implies the ordinary expectation error bound
The conditional version is stronger in that it bounds the entire random difference between conditional expectations in \(L^1\), not just the difference between their overall means. This is often useful when an input variable is replaced by a tractable approximation before conditioning.
A common mistake is to read the contraction inequality as a pointwise comparison between \(|\mathbb{E}[X\mid\mathcal{G}]|\) and \(|X|\). The theorem compares expectations of absolute values; it does not assert that the conditional average is smaller than \(|X|\) at each outcome. The pointwise bound supplied by the lemma is instead against \(\mathbb{E}[|X|\mid\mathcal{G}]\). Keeping these two statements distinct is essential.
Check Your Understanding
Use the definitions and inequalities in this tutorial to answer the following questions.
- State the \(L^1\) contraction inequality for an integrable random variable \(X\) and a sigma-algebra \(\mathcal{G}\).
- Why does the conditional absolute-value bound imply the \(L^1\) contraction after taking expectations?
- If \(\|X-Y\|_1\leq\delta\), what bound follows for the \(L^1\) distance between their conditional expectations given the same sigma-algebra?
- When \(\mathcal{G}\subseteq\mathcal{H}\), which residual controls the \(L^1\) difference between the two conditional averages?
- Does the contraction theorem imply \(|\mathbb{E}[X\mid\mathcal{G}]|\leq|X|\) almost surely? Explain the distinction.