Conditional Absolute Values and Bounds

Why Conditional Bounds Matter

A conditional expectation summarizes an integrable random variable using the information in a sigma-algebra. Bounds on the original variable therefore give useful control over its conditional expectation. The central absolute-value inequality says that taking a conditional expectation cannot make the absolute value larger than the conditional expectation of the absolute value.

This inequality is also a practical starting point for transferring more general bounds. If a random variable is almost surely between two integrable random variables, its conditional expectation is almost surely between their conditional expectations. When the bounds are already measurable with respect to the conditioning sigma-algebra, they pass through unchanged.

Setup and Notation

Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space, let \(\mathcal G\subseteq\mathcal F\) be a sigma-algebra, and let \(X\) be a real-valued integrable random variable. A conditional expectation \(\mathbb E[X\mid\mathcal G]\) is a \(\mathcal G\)-measurable integrable random variable whose integral over every \(A\in\mathcal G\) equals the integral of \(X\) over \(A\). It is determined only up to almost-sure equality.

We use two standard properties established in the earlier material on conditional expectation. Linearity allows sums and scalar multiples to pass through conditional expectation. Order preservation says that if \(X\le Y\) almost surely for integrable random variables \(X,Y\), then \(\mathbb E[X\mid\mathcal G]\le\mathbb E[Y\mid\mathcal G]\) almost surely.

Definition: An almost-sure bound \(X\le Y\) means that \(\mathbb P(X\le Y)=1\). It need not hold at every point of \(\Omega\). All inequalities between conditional expectations below are understood almost surely.

The Conditional Absolute-Value Inequality

For an integrable real random variable \(X\), the absolute value \(|X|\) is integrable as well. Thus both \(\mathbb E[X\mid\mathcal G]\) and \(\mathbb E[|X|\mid\mathcal G]\) are defined. The latter is nonnegative almost surely by order preservation.

Theorem: If \(X\) is a real-valued integrable random variable and \(\mathcal G\subseteq\mathcal F\) is a sigma-algebra, then \[ \left|\mathbb E[X\mid\mathcal G]\right| \le \mathbb E[|X|\mid\mathcal G] \quad\text{almost surely}. \]

Proof. The pointwise inequalities \(-|X|\le X\le |X|\) hold everywhere. Applying order preservation to both inequalities gives \[ \mathbb E[-|X|\mid\mathcal G] \le \mathbb E[X\mid\mathcal G] \le \mathbb E[|X|\mid\mathcal G] \quad\text{almost surely}. \] By linearity, \(\mathbb E[-|X|\mid\mathcal G]=-\mathbb E[|X|\mid\mathcal G]\). Hence the conditional expectation of \(X\) lies between \(-\mathbb E[|X|\mid\mathcal G]\) and \(\mathbb E[|X|\mid\mathcal G]\). Since the latter is nonnegative, this is exactly the claimed absolute-value inequality. \(\square\)

The result compares two random variables, both measurable with respect to \(\mathcal G\). It does not say that the conditional expectation of \(|X|\) is a fixed numerical bound; it can vary with the information represented by \(\mathcal G\).

Worked Example: Checking the Inequality on a Finite Partition

Let \(\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\}\), with each outcome having probability \(1/4\). Let \(\mathcal G\) be generated by the partition \(A=\{\omega_1,\omega_2\}\), \(B=\{\omega_3,\omega_4\}\), and define \(X\) by the values \(2,-4,1,3\), respectively.

On each cell, conditional expectation is the average of \(X\) on that cell. Therefore \(\mathbb E[X\mid\mathcal G]\) equals \((2+(-4))/2=-1\) on \(A\), and \((1+3)/2=2\) on \(B\). Thus its absolute value equals \(1\) on \(A\) and \(2\) on \(B\).

For \(|X|\), the values are \(2,4,1,3\). Its conditional averages are \((2+4)/2=3\) on \(A\) and \((1+3)/2=2\) on \(B\). The asserted comparison is therefore \(1\le3\) on \(A\) and \(2\le2\) on \(B\), as required.

Transferring Almost-Sure Bounds

The absolute-value inequality is one instance of a broader consequence of order preservation. The bounds may themselves be random, provided they are integrable.

Theorem: Let \(L,X,U\) be integrable real-valued random variables. If \(L\le X\le U\) almost surely, then \[ \mathbb E[L\mid\mathcal G] \le \mathbb E[X\mid\mathcal G] \le \mathbb E[U\mid\mathcal G] \quad\text{almost surely}. \]

Proof. From \(L\le X\) almost surely, order preservation gives \(\mathbb E[L\mid\mathcal G]\le\mathbb E[X\mid\mathcal G]\) almost surely. From \(X\le U\) almost surely, it gives \(\mathbb E[X\mid\mathcal G]\le\mathbb E[U\mid\mathcal G]\) almost surely. The two inequalities hold together outside the union of two null sets, which is still a null set. This proves the claimed almost-sure sandwich. \(\square\)

In particular, if \(m\) and \(M\) are constants with \(m\le X\le M\) almost surely, then \[ m\le\mathbb E[X\mid\mathcal G]\le M \quad\text{almost surely}. \] Here the constants pass through conditional expectation because a constant is \(\mathcal G\)-measurable and integrable. More generally, if an integrable random variable \(Z\) is \(\mathcal G\)-measurable, then \(Z\) itself is a version of \(\mathbb E[Z\mid\mathcal G]\): it is \(\mathcal G\)-measurable and has the required integrals over every event in \(\mathcal G\). Consequently, \(\mathcal G\)-measurable bounds pass through unchanged.

Worked Example: Deterministic Bounds Remain Valid

On four equally likely outcomes, let \(X\) take the values \(-2,-1,3,4\), and take \(\mathcal G=\{\varnothing,\Omega\}\), the trivial sigma-algebra. The almost-sure bounds are \(-2\le X\le4\).

Conditioning on the trivial sigma-algebra gives the constant mean: \[ \mathbb E[X\mid\mathcal G] = \frac{-2+(-1)+3+4}{4} =1. \] Thus the transferred bounds are \(-2\le1\le4\), which hold. The absolute-value inequality also gives a valid, though less sharp, estimate: \(\mathbb E[|X|\mid\mathcal G]=(2+1+3+4)/4=5/2\), so \(|1|\le5/2\).

Bounds Measurable with Respect to the Conditioning Information

A useful special case occurs when the bounds are already known given \(\mathcal G\). If \(L\) and \(U\) are integrable and \(\mathcal G\)-measurable, and \(L\le X\le U\) almost surely, then \[ L\le\mathbb E[X\mid\mathcal G]\le U \quad\text{almost surely}. \] Indeed, the general sandwich result applies, and \(\mathbb E[L\mid\mathcal G]=L\) and \(\mathbb E[U\mid\mathcal G]=U\) almost surely.

A centered version is often convenient. Let \(Z\) be an integrable \(\mathcal G\)-measurable random variable, and let \(Y\) be integrable with \(|X-Z|\le Y\) almost surely. This assumption itself implies \(Y\ge0\) almost surely. Linearity gives \(\mathbb E[X\mid\mathcal G]-Z=\mathbb E[X-Z\mid\mathcal G]\), and the conditional absolute-value inequality then yields \[ \left|\mathbb E[X\mid\mathcal G]-Z\right| \le \mathbb E[|X-Z|\mid\mathcal G] \le \mathbb E[Y\mid\mathcal G] \quad\text{almost surely}. \] The last inequality follows from order preservation. This form controls how far the conditional expectation can be from a \(\mathcal G\)-measurable center.

Worked Example: Random Bounds on Partition Cells

Let \(\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\}\) have equal probabilities, and let \(\mathcal G\) be generated by \(A=\{\omega_1,\omega_2\}\) and \(B=\{\omega_3,\omega_4\}\). Set \(X=(1,3,-2,4)\) on the four outcomes. Define the \(\mathcal G\)-measurable variables \(L\) and \(U\) to be \(0\) and \(4\) on \(A\), and \(-3\) and \(5\) on \(B\), respectively.

The bounds hold outcome by outcome: on \(A\), both \(1\) and \(3\) lie in \([0,4]\); on \(B\), both \(-2\) and \(4\) lie in \([-3,5]\). The conditional averages of \(X\) are \((1+3)/2=2\) on \(A\) and \((-2+4)/2=1\) on \(B\). Hence \(L\le\mathbb E[X\mid\mathcal G]\le U\) reads \(0\le2\le4\) on \(A\) and \(-3\le1\le5\) on \(B\).

Common Pitfalls and Uses

The inequalities are almost-sure statements, not assertions about every point of the sample space. Conditional expectations are only specified up to almost-sure equality, so changing a version on a null set cannot affect the conclusion. If several inequalities are combined, their exceptional null sets can be combined as well: a finite union of null sets is null.

Integrability is also essential to the stated results. The assumptions ensure that the conditional expectations appearing in the inequalities are defined as integrable random variables. A bound on \(X\) that does not ensure integrability does not by itself justify writing these conditional expectations.

The absolute-value inequality is a useful way to turn control of \(X\) into control of its conditional expectation. For example, if \(|X|\le Y\) almost surely for an integrable nonnegative \(Y\), order preservation gives \(\mathbb E[|X|\mid\mathcal G]\le\mathbb E[Y\mid\mathcal G]\). Combining this with the main inequality yields \(|\mathbb E[X\mid\mathcal G]|\le\mathbb E[Y\mid\mathcal G]\) almost surely. The result gives pointwise conditional control without requiring \(X\) to be \(\mathcal G\)-measurable.

Key takeaway: Order preservation transfers almost-sure bounds through conditional expectation. Applying it to \(-|X|\le X\le|X|\) gives the conditional absolute-value inequality.

Check Your Understanding

Use the assumptions carefully, and interpret each conclusion as an almost-sure statement.

Which two pointwise inequalities for \(X\) lead directly to the conditional absolute-value inequality?
If \(L\le X\le U\) almost surely, what bounds hold for \(\mathbb E[X\mid\mathcal G]\)?
Why does a \(\mathcal G\)-measurable integrable bound pass through conditional expectation unchanged?
If \(Z\) is \(\mathcal G\)-measurable and \(|X-Z|\le Y\) almost surely, what conditional bound controls \(\left|\mathbb E[X\mid\mathcal G]-Z\right|\)?
Why should inequalities between versions of conditional expectations be understood almost surely rather than at every sample point?