Conditional Expectation as an Lp Contraction

Why Conditioning Reduces an \(L^p\) Norm

Conditional expectation replaces a random variable by the information visible to a chosen sigma-algebra. In doing so, it averages away some variation. The \(L^p\) contraction theorem makes a precise version of this idea: for every \(p\) between \(1\) and \(\infty\), conditioning cannot increase the \(L^p\) norm.

This bound is useful whenever conditional expectations are treated as operators on spaces of random variables. It guarantees that an input with finite \(p\)-th moment still has a conditional expectation with finite \(p\)-th moment, and controls its size uniformly. The result includes both the \(L^1\) contraction and the essential-supremum bound at \(p=\infty\).

Spaces and the Contraction Statement

Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space and let \(\mathcal G\subseteq\mathcal F\) be a sigma-algebra. For \(1\leq p<\infty\), \(L^p\) consists of equivalence classes of measurable random variables \(X\) for which \(\mathbb E[|X|^p]<\infty\), with norm \(\|X\|_p=(\mathbb E[|X|^p])^{1/p}\). The space \(L^\infty\) consists of essentially bounded random variables, with norm \(\|X\|_\infty=\operatorname*{ess\,sup}|X|\).

For \(X\in L^1\), the conditional expectation \(\mathbb E[X\mid\mathcal G]\) is a \(\mathcal G\)-measurable random variable whose integral over every event in \(\mathcal G\) agrees with that of \(X\). It is defined up to almost-sure equality. Since the probability space has total mass one, \(L^p\subseteq L^1\) whenever \(1\leq p<\infty\), so this conditional expectation is defined for every finite-\(p\) input considered below.

The Conditional Jensen Inequality

For \(p>1\), the key estimate is a conditional version of Jensen’s inequality: \[ |\mathbb E[X\mid\mathcal G]|^p \leq \mathbb E[|X|^p\mid\mathcal G] \quad\text{almost surely}. \] Here it is enough to prove this particular inequality. The proof uses supporting affine functions of the convex function \(x\mapsto |x|^p\), and only the positivity and linearity properties of conditional expectation.

For each rational number \(a\), define \[ \ell_a(y)=|a|^p+p\,\operatorname{sgn}(a)|a|^{p-1}(y-a), \] where the coefficient of \(y-a\) is taken to be \(0\) when \(a=0\). Convexity and differentiability of \(y\mapsto |y|^p\) imply \(\ell_a(y)\leq |y|^p\) for every real \(y\), with equality when \(y=a\). Moreover, for every real \(z\), \[ |z|^p=\sup_{a\in\mathbb Q}\ell_a(z). \] Indeed, each term in the supremum is at most \(|z|^p\); choosing rational numbers \(a\) tending to \(z\), continuity of the function and its derivative shows that \(\ell_a(z)\) tends to \(|z|^p\).

Proof of the conditional Jensen inequality. Let \(X\in L^p\) with \(p>1\), and write \(Y=\mathbb E[X\mid\mathcal G]\). For each rational \(a\), the pointwise inequality \(\ell_a(X)\leq |X|^p\) can be conditioned. The affine form of \(\ell_a\), together with linearity, gives \[ \ell_a(Y)\leq\mathbb E[|X|^p\mid\mathcal G]\quad\text{almost surely}. \] All the quantities here are integrable: \(X\in L^1\), and \(|X|^p\in L^1\). There are only countably many rational \(a\), so the inequalities hold simultaneously outside one null set. Taking their supremum and using the representation above yields \[ |Y|^p=\sup_{a\in\mathbb Q}\ell_a(Y) \leq\mathbb E[|X|^p\mid\mathcal G]\quad\text{almost surely}. \]

Proof of the \(L^p\) Contraction Theorem

Proof. First suppose \(p=1\). The inequalities \(-|X|\leq X\leq |X|\), together with order preservation and linearity of conditional expectation, give \[ -\,\mathbb E[|X|\mid\mathcal G] \leq \mathbb E[X\mid\mathcal G] \leq \mathbb E[|X|\mid\mathcal G]. \] Thus \(|\mathbb E[X\mid\mathcal G]|\leq\mathbb E[|X|\mid\mathcal G]\) almost surely. Taking expectations and using \(\mathbb E[\mathbb E[|X|\mid\mathcal G]]=\mathbb E[|X|]\) proves the claim for \(p=1\). Now let \(1<p<\infty\). The conditional Jensen inequality just proved and the defining integral property of conditional expectation imply \[ \mathbb E[|\mathbb E[X\mid\mathcal G]|^p] \leq \mathbb E[\mathbb E[|X|^p\mid\mathcal G]] = \mathbb E[|X|^p]. \] In particular, the conditional expectation belongs to \(L^p\). Taking \(p\)-th roots proves the norm bound. Finally, let \(X\in L^\infty\) and set \(M=\|X\|_\infty\). For every \(\varepsilon>0\), \(|X|\leq M+\varepsilon\) almost surely. Order preservation and the fact that conditional expectation preserves constants give \[ -(M+\varepsilon) \leq\mathbb E[X\mid\mathcal G] \leq M+\varepsilon \quad\text{almost surely}. \] Taking \(\varepsilon\) through the sequence \(1,1/2,1/3,\ldots\) shows that \(|\mathbb E[X\mid\mathcal G]|\leq M\) almost surely. Hence the conditional expectation is essentially bounded and its \(L^\infty\) norm is at most \(M\). \(\square\)

Worked Examples

Worked Example: Conditioning on a Finite Partition

Let \(\Omega=\{1,2,3,4\}\), give each point probability \(1/4\), and let \(\mathcal G\) be generated by the partition \(\{\{1,2\},\{3,4\}\}\). Set \(X(1)=1\), \(X(2)=3\), \(X(3)=-2\), and \(X(4)=0\). On each block, the conditional expectation is the average of the values there. Therefore \(Y=\mathbb E[X\mid\mathcal G]\) takes value \(2\) on \(\{1,2\}\) and \(-1\) on \(\{3,4\}\).

For every finite \(p\geq1\), \[ \|X\|_p^p=\frac{1+3^p+2^p}{4}, \qquad \|Y\|_p^p=\frac{2^p+1}{2}. \] For example, at \(p=2\), these are \(14/4=7/2\) and \(5/2\), respectively. At \(p=1\), both equal \(3/2\). The example shows both that the norm can decrease and that equality can occur even when \(X\) is not \(\mathcal G\)-measurable.

Worked Example: The \(L^\infty\) Bound

On the same four-point probability space, let \(\mathcal G\) be generated by \(\{\{1,2\},\{3,4\}\}\), and take \(X\) to have values \(4,-2,1,3\) at \(1,2,3,4\), respectively. Its essential supremum norm is \(\|X\|_\infty=4\). The conditional averages are \(1\) on the first block and \(2\) on the second, so \[ \|\mathbb E[X\mid\mathcal G]\|_\infty=2\leq4=\|X\|_\infty. \] The conditional expectation is still bounded, and averaging can lower the largest absolute value.

Worked Example: Equality for Every \(p\)

Let \(\Omega=\{1,2,3,4\}\) have the uniform probability measure, and let \(\mathcal G\) be generated by \(\{\{1,2\},\{3,4\}\}\). Define \(X\) to equal \(1\) on \(\{1,2\}\) and \(-1\) on \(\{3,4\}\). This \(X\) is \(\mathcal G\)-measurable, so its conditional expectation is \(X\) itself. For finite \(p\geq1\), \(\|X\|_p^p=\mathbb E[|X|^p]=1\), and \(\|X\|_\infty=1\). Thus equality holds in the contraction bound for every \(p\) in the theorem.

Sharpness and Equality

The constant one in the contraction inequality cannot be improved. In fact, for any probability space and any sub-sigma-algebra \(\mathcal G\), conditional expectation has operator norm exactly one on each \(L^p\), \(1\leq p\leq\infty\). The contraction theorem gives the upper bound. For the lower bound, use the constant random variable \(X=1\): its conditional expectation is also \(1\), and both norms are one.

Proposition: As an operator from \(L^p\) to \(L^p\), conditional expectation has norm one for each \(1\leq p\leq\infty\).

Proof. The contraction theorem implies that the operator norm is at most one. The constant random variable \(1\) belongs to every \(L^p\), has norm \(\|1\|_p=1\), and satisfies \(\mathbb E[1\mid\mathcal G]=1\). Therefore the ratio of output norm to input norm is one, so the operator norm is at least one. Combining the bounds gives exactly one. \(\square\)

Equality in the norm inequality does not, in general, imply that \(X\) is \(\mathcal G\)-measurable: the first worked example already gives equality at \(p=1\) for a variable that changes within each partition block. A safe equality example for every \(p\) is a \(\mathcal G\)-measurable input, whose conditional expectation equals that input almost surely. The contraction theorem itself needs no characterization of all equality cases.

Why the Endpoint Cases Matter

The finite-\(p\) proof has two different mechanisms at its endpoints. At \(p=1\), the key estimate is the conditional absolute-value inequality, followed by integration. For \(p>1\), convexity supplies the conditional Jensen inequality. At \(p=\infty\), there is no finite moment to integrate; instead, an almost-sure bound on the input is preserved by order preservation and the preservation of constants.

A common mistake is to infer an \(L^\infty\) bound merely from finiteness of some \(L^p\) norm. A finite \(p\)-th moment does not imply essential boundedness. The endpoint argument specifically assumes \(X\in L^\infty\). Conversely, for finite \(p\), the theorem proves that the conditional expectation belongs to \(L^p\); this is a conclusion, not an assumption about its integrability.

Check Your Understanding

Try to answer these questions using the definitions and arguments above.

Why is \(\mathbb E[X\mid\mathcal G]\) defined when \(X\in L^p\) for finite \(p\geq1\) on a probability space?
How does the conditional Jensen inequality yield the \(L^p\) contraction for \(1<p<\infty\)?
Which order bounds establish the \(L^1\) estimate?
Why does the \(L^\infty\) proof use \(M+\varepsilon\) before concluding a bound by \(M\)?
What input demonstrates that the operator norm is at least one?