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Conditional Expectation · Tutorial 228 of 946

Conditioning on a Sigma-Algebra

Learn how a conditional expectation is defined by the information available and by agreement of integrals on every event that information can describe.

Intermediate 9 min read

What You'll Learn

  • State the definition of conditional expectation relative to a sub-sigma-algebra.
  • Explain why measurability encodes dependence only on specified information.
  • Interpret the integral-matching condition as preservation of averages on observable events.
  • Verify conditional expectations in cases of no information, complete information, and a coarse observation.
  • Distinguish conditional expectation given a sigma-algebra from a pointwise conditional value.

Introduction

An expectation summarizes a random variable by a single number. But in many problems, some information is available and some is not. We may want a random variable that represents the expected value of \(X\) using only the events that the available information can distinguish.

A sigma-algebra provides a precise way to specify that information. Conditional expectation relative to a sigma-algebra is defined by two requirements: the proposed answer must be measurable with respect to that sigma-algebra, and its integrals must agree with those of the original random variable over every event in the sigma-algebra.

The Definition

Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space, let \(\mathcal G\subseteq\mathcal F\) be a sub-sigma-algebra, and let \(X\) be a real-valued integrable random variable. Integrability means \(\mathbb E[|X|]<\infty\). The sigma-algebra \(\mathcal G\) represents the information on which the conditional expectation is allowed to depend.

Definition: A conditional expectation of \(X\) given \(\mathcal G\) is an integrable, \(\mathcal G\)-measurable random variable \(Y\) such that, for every event \(A\in\mathcal G\), $$ \int_A Y\,d\mathbb P=\int_A X\,d\mathbb P. $$ Such a random variable is denoted by \(\mathbb E[X\mid\mathcal G]\).

The definition imposes two different kinds of constraint. Measurability restricts what information \(Y\) may use. The integral identity requires \(Y\) to preserve the average of \(X\) on each event that can be described using that information.

The quantifier over \(A\in\mathcal G\) is essential. Agreement only on the whole sample space would require the two variables to have the same overall expectation, but would not ensure that their averages agree on the events visible to \(\mathcal G\).

What the Two Requirements Mean

A random variable \(Y\) is \(\mathcal G\)-measurable when its values can be determined using the information encoded by \(\mathcal G\). In particular, if an event about the value of \(Y\) occurs, that event must belong to \(\mathcal G\). Thus \(Y\) cannot distinguish outcomes that the available information itself cannot distinguish in a measurable way.

The integral condition can be read as a balancing requirement. For each observable event \(A\), the total contribution of \(Y\) on \(A\) must equal the total contribution of \(X\) on \(A\). If \(\mathbb P(A)>0\), division by \(\mathbb P(A)\) shows that the averages over \(A\) agree:

$$ \frac{1}{\mathbb P(A)}\int_A Y\,d\mathbb P = \frac{1}{\mathbb P(A)}\int_A X\,d\mathbb P. $$

This does not say that \(Y\) equals \(X\) at each outcome in \(A\). It says that \(Y\) represents \(X\)'s average contribution on every event that the information can identify.

Worked Examples

Worked Example: No Information

Take the trivial sigma-algebra \(\mathcal G=\{\varnothing,\Omega\}\), and let \(X\) be any integrable random variable. The constant \(Y=\mathbb E[X]\) is \(\mathcal G\)-measurable and integrable. For \(A=\varnothing\), both integrals are zero. For \(A=\Omega\),

$$ \int_\Omega Y\,d\mathbb P =\mathbb E[X]\mathbb P(\Omega) =\mathbb E[X] =\int_\Omega X\,d\mathbb P. $$

These are the only two events in \(\mathcal G\), so \(Y\) satisfies the definition. With no event distinctions available, the conditional expectation is represented by the overall mean.

Worked Example: Complete Information

Suppose \(\mathcal G=\mathcal F\). Then \(X\) itself is \(\mathcal G\)-measurable and integrable. For every \(A\in\mathcal G\), the required identity is simply

$$ \int_A X\,d\mathbb P=\int_A X\,d\mathbb P. $$

Therefore \(X\) is a conditional expectation of itself given the full sigma-algebra. When all the events in the underlying sigma-algebra are available, no information restriction prevents using \(X\).

Worked Example: Observing Which Half of the Interval Contains the Outcome

Let \(\Omega=[0,1]\) with Lebesgue probability measure, let \(X(t)=t\), and let \(A=[0,\tfrac12]\). Consider the information \(\mathcal G=\{\varnothing,A,A^c,\Omega\}\), which records which half contains the outcome. Define \(Y(t)=\tfrac14\) on \(A\) and \(Y(t)=\tfrac34\) on \(A^c\). This variable is \(\mathcal G\)-measurable and integrable.

On the first half,

$$ \int_A X\,d\mathbb P=\int_0^{1/2}t\,dt=\frac18, \qquad \int_A Y\,d\mathbb P=\frac14\cdot\frac12=\frac18. $$

On the second half,

$$ \int_{A^c} X\,d\mathbb P=\int_{1/2}^{1}t\,dt=\frac38, \qquad \int_{A^c}Y\,d\mathbb P=\frac34\cdot\frac12=\frac38. $$

For \(\Omega\), the integrals are the sums of these values, both equal to \(\tfrac12\); for \(\varnothing\), both are zero. These are all the events in \(\mathcal G\). Thus \(Y\) satisfies the defining integral identities. It retains the average of \(X\) on each half, without using the exact value of the outcome.

Basic Consequences of the Definition

Theorem: If \(X\) is integrable, then the constant \(\mathbb E[X]\) is a conditional expectation of \(X\) given the trivial sigma-algebra \(\{\varnothing,\Omega\}\).

Proof. The constant \(Y=\mathbb E[X]\) is integrable and measurable with respect to the trivial sigma-algebra. Its integral over \(\varnothing\) is zero, as is the integral of \(X\). Its integral over \(\Omega\) is

$$ \int_\Omega Y\,d\mathbb P =\mathbb E[X]\mathbb P(\Omega) =\mathbb E[X] =\int_\Omega X\,d\mathbb P. $$

Since \(\varnothing\) and \(\Omega\) are the only events in the trivial sigma-algebra, the defining identity holds for every such event. Therefore \(Y\) is a conditional expectation.

\(\square\)

Theorem: If \(X\) is integrable and \(\mathcal G\)-measurable, then \(X\) is a conditional expectation of \(X\) given \(\mathcal G\).

Proof. By assumption, \(X\) is \(\mathcal G\)-measurable, and it is integrable. For every \(A\in\mathcal G\), setting \(Y=X\) gives

$$ \int_A Y\,d\mathbb P=\int_A X\,d\mathbb P. $$

Thus both requirements in the definition hold, so \(X\) is a conditional expectation of itself given \(\mathcal G\).

\(\square\)

Proposition: Any conditional expectation \(Y\) of an integrable random variable \(X\) given \(\mathcal G\) has the same overall expectation as \(X\).

Proof. The whole sample space \(\Omega\) belongs to every sigma-algebra. Applying the defining identity with \(A=\Omega\) gives

$$ \mathbb E[Y] =\int_\Omega Y\,d\mathbb P =\int_\Omega X\,d\mathbb P =\mathbb E[X]. $$

This proves the assertion.

\(\square\)

Information, Averages, and Common Pitfalls

The phrase “given \(\mathcal G\)” refers to a collection of observable events, not necessarily to a single event. The conditional expectation is a random variable because the information may vary across outcomes: different observations can lead to different values of the conditional expectation.

It is also important not to treat \(\mathbb E[X\mid\mathcal G]\) as a pointwise answer obtained by conditioning on each individual outcome. The definition is expressed through integrals over events in \(\mathcal G\). In general, it does not assert that the conditional expectation equals \(X\) at any particular outcome, or that its value is an average over a set consisting of that outcome.

A second common pitfall is to check only the total expectation. The identity on \(\Omega\) is necessary, but the definition asks for agreement on every \(A\in\mathcal G\). The interval example shows why those additional tests matter: the candidate matches \(X\)'s average on each observable half, not merely across the entire interval.

The definition tells us what a conditional expectation must satisfy; by itself, it is not a construction. The Radon–Nikodym theorem provides the standard existence argument for integrable \(X\) and a sub-sigma-algebra \(\mathcal G\). The key conceptual point here is that the constructed random variable must obey both constraints: it uses only the specified information and preserves integrals on every event that information describes.

Check Your Understanding

Use the definition to explain what information is allowed and what the integral identities enforce.

  1. What are the two requirements in the definition of \(\mathbb E[X\mid\mathcal G]\)?
  2. Why is the integral identity over \(\Omega\) alone not the full definition?
  3. What does the trivial sigma-algebra represent, and why does its conditional expectation reduce to a constant mean?
  4. If \(X\) is already \(\mathcal G\)-measurable, why does \(X\) satisfy the definition?
  5. In the interval example, verify the identity on \(A^c\) and explain what the value \(\tfrac34\) represents.

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